Yearly Archives: 2021

NCERT Exercises

Question 1.
A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.4 A. What is the magnitude of the magnetic field 6 at the centre of the coil?
Solution:
The magnetic field at the centre of a circular coil having 100 turns.

Question 2.
A long straight wire carries a current of 35 A. What is the magnitude of the field 6 at a point 20 cm from the wire?
Solution:
Magnetic field due to a long straight wire

Question 3.
A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.
Solution:
Let us first decide the standard directions on the plane of paper.

Magnitude of magnetic field

Question 4.
A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?
Solution:
The standard directions on the plane of paper can be different according to requirements.

Direction of magnetic field can be observed by right hand palm rule and it is southward.

Question 5.
What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?
Solution:

Let l be length of wire, carrying a current of 8 A at an angle 30° with the magnetic field. Force on the wire, F = |B| sinθ Force per unit length F/l = |B| sinθ F/l = 8 × 0.15 × sin 30° = 0.6 N m^-1.

Question 6.
A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?
Solution:
The magnetic field inside the solenoid is along its axis. the current in the wire flows perpendicular to the axis. F = |B| sin 90° or F = 10 × 0.27 × 3 × 10^-2 x 1 = 0.081 N

Question 7.
Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
Solution:
Force of attraction per unit length on two parallel wires carrying current in same direction.

Question 8.
A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current the carried is 8.0 A, estimate the magnitude of 8 inside the solenoid near its centre.
Solution:
Total number of turns in 80 cm length of solenoid can be calculated

Question 9.
A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
Solution:

Torque experienced by the coil carrying current in the given magnetic field.

Question 10.
Two moving coil meters, M₁ and M₂ have following particulars: R₁ = 10Ω, N₁ = 30, A₁ = 3.6 × 10^-3 m², B₁ = 0.25 T R₂ = 14 Ω, N₂ = 42, A₂ = 1.8 × 10^-3 m², B₂ = 0.50 T. (The spring constants are identical for two meters). Determine ratio of (i) current sensitivity (ii) voltage sensitivity of M2 and A4,.
Solution:
Current sensitivity of a moving coil galvanometer is defined as

(ii) Ratio of voltage sensitivity

Question 11.
In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^-4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m s^-1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e= 1.6 × 10^-19 C,m_e = 9.1 × 10^-31 kg)
Solution:
The magnetic force ƒ = quB act normal to the direction of motion, thus provide the necessary centripetal force to follow the circular path.

Question 12.
In question 11, obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
Solution:
Time period can be calculated.

The frequency of revolution of electron is independent of speed of electron.

Question 13.
(a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? [All other particulars are also unaltered].
Solution:
(a) The given coil is circular and is suspended such that field lines makes angle 60° with normal of the coil.A similar torque is required to prevent the coil from turning.

A similar torque is required to prevent the coil from turning.
(b) As long as the area of the planar coil remains same, the torque on the coil is also same, irrespective of the shape.

Question 14.
Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.
Solution:
The concentric coils are in the plane north to south. Let us decide the directions with conveniences in the plane of paper.

Magnetic field at centre due to coil X

So, a magnetic field nearly 1.6 × 10^-3 T will appear at centre of the circular coil directed towards west.

Question 15.
A magnetic field of 100 G (1G = 10^-4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10^3-3 m². The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m”1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.
Solution:
The magnetic field required is 100 G in a length of 10 cm. So, the solenoid should have a length larger than 10 cm. The maximum current capacity of wire is 15 A, so less than 15 A should flow in wire. Now using

So, one combination for the desired 100 G magnetic field in a length 10 cm can be a total length of solenoid of 50 cm and current in the solenoid of 10 A and total turns nearly 400.

Question 16.
Fora circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its

(a) Show that this reduces to the familiar result for field at the centre of the coil
(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,

[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]
Solution:
On the axis of the circular coil of radius R and N turns carrying a current I at a distance x from centre the magnetic field is

Which is the same result as is obtained by integration at centre of the coil.
(b) The side of the loop where current flows clockwise becomes south pole and where the current appear anticlockwise becomes north pole. Direction of magnetic field is from south pole to north pole.

Now, two parallel coils each of radius R with N turns. The magnetic field of both the coils add.

Question 17.
A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field
(a) outside the toroid,
(b) inside the core of the toroid, and
(c) in the empty space surrounded by the toroid.
Solution:

Total turns in the toroid are given 3500.
(a) outside the toroid the magnetic field is zero.
(b) Inside the core of toroid, the magnetic field will be

(c) Magnetic field in the empty space surrounded by the toroid is also zero.

Question 18.
Answer the following questions:
(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?
(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?
(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.
Solution:
(a) If a charged particle move parallel or anti parallel to the magnetic field, no magnetic force will act on it and it move undeflected. So, in the given condition either the charged particle enters east to west or west to east.
(b) A magnetic force can only change the direction of charged particle but never changes magnitude of speed as force act normal to direction of speed. So, charged particle may follow a complicated trajectory, but its speed remains the same.
(c)

If,we want the electron to move undeflected in the presence of electric and magnetic fields, then the electric force should be balanced by magnetic force.

So, the magnetic field should act in downward direction.

Question 19.
An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30° with the initial velocity.
Solution:
A electron which is accelerated by a potential difference of 2.0 kV will have a kinetic energy gained 2000 eV.

(a) When the electron enters in the uniform r magnetic field which is normal to the velocity of electron the electron follows a circular path r of radius.

(b) When the magnetic field makes an angle 30° with the initial velocity, the trajectory of the electron becomes helical. radius of the helical path is

Question 20.
A magnetic field set up using Helmholtz coils (described in question 16 above) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 × 10^-5 V m^-1, make a simple guess as to what the beam contains. Why is the answer not unique?
Solution:
Narrow beam of charged particles remains undeflected and is perpendicular to both electric field and magnetic fields which are mutually perpendicular. So, the electric force is balanced by magnetic force. qE = quB Speed of charged particles

Because the beam is accelerated through 15 kV, if charge is q, then kinetic energy gained by charged particles

Here, we can only obtain charge to mass ratio and same ratio can be in Deuterium ions, He⁺⁺, Li⁺⁺⁺, so the beam can contain any of these charged particles.

Question 21.
A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before?
[Ignore the mass of the wires.] g = 9.8 m s^-2
Solution:

Tension in the strings and magnetic force |B| balance the weight of wire. 2 T + |B| = mg

Question 22.
The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?
Solution:
Two wires connecting battery of an automobile to the starting motor carry 300 A current in opposite direction. So, the force is repulsive between them. Force per unit length

Question 23.
A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,
(a) the wire intersects the axis,
(b) the wire is turned from N-S to northeast- northwest direction.
(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?
Solution:

The magnetic field is in the direction east to west and in the cylindrical region of radius 10 cm.
(a) A current carrying wire, intersects the axis. Force on wire, F = |B| = IB (2r) = 7 × 1.5 × 2 × 1 × 10^-2 F = 2.1 N Downwards
(b) By turning the wire by an angle 45° in NE and NW direction the force remain the same,/= 2.1 N Downwards
(c) Now the wire is lowered from the axis by a distance of 6.0 cm.

Length of wire inside cylindrical region is 16 cm now. So, the force is F = |B| = 7 × 1.5 × 16 × 10^-2 = 1.68 N, vertically downwards

Question 24.
A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in figure? What is the force on each case? Which case corresponds to stable equilibrium?

Solution:
(a) Let us detail each case separately,


So, torque is 1.8 × 10^-2 N-m along -y direction net force on the coil is zero coil not in equilibrium.
(b) Dipole moment is along +x direction

So, torque is 1.8 x 10-2 N m along -y direction. Net force on the coil is zero, coil is not in equilibrium
(c) Dipole moment is along -y direction

Net force on the coil is zero, coil is not in equilibrium.
(d)

Dipole moment is at an angle 150° with the +x direction.

At an angle 240° with the +x direction net force on the coil is zero, coil is not in equilibrium.

Negative energy shows equilibrium is stable,
(f) Dipole moment is along – z direction

Positive energy shows equilibrium is unstable.

Question 25.
A circular coil of 20 turns and radius 10 cm is place in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross-sectional area 10^-5 m², the free electron density in copper is given to be about 10²⁹   m-³.)
Solution:
The magnetic field is normal to the plane of the coil, so condition of minimum torque.

(a) Torque on the coil τ = NIB A sinθ here θ = 0° τ = 0
(b) Force on every element of the coil is cancelled by force on corresponding element. Net force on the unit is zero.

(c) To calculate force on each electron, let us find drift velocity.

Question 26.
A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside solenoid (near its centre) normal to its axis; both the wire and axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current [with appropriate sense of circulation] in the windings of the solenoid can support the weight of the wire? g = 9.8 m s^-2.
Solution:
Magnetic in the solenoid B = μ_0nI

A current carrying wire is suspended inside solenoid and a current 7 = 6.0 A is flowing in it. To balance the weight of the wire by the magnetic force

Question 27.
A galvanometer coil has a resistance of 12 Q and the metre shows full scale deflection for a current of 3 mA. How will you convert metre into a voltmeter of range 0 to 18 V?
Solution:
By using the formula

We can calculate required resistance to be connected in series.

Question 28.
A galvanometer coil has a resistance of 15 Q and the meter shows full scale deflection for a current of 4 mA. How will you convert metre into an ammeter of range 0 to 6 A?
Solution:
For converting galvanometer into ammeter of required range, required shunt can be calculated

NCERT Exercises

Question 1.
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?
Solution:
The maximum current will be obtained when no external resistance is offered by wire joining the two terminals.

Question 2.
A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Solution:

Question 3.
(a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Solution:
(a) Total resistance of the combination in series

Question 4.
(a) Three resistors 2Ω, 4Ω and 5Ω are combined in parallel. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance. Determine the current through each resistor, and the total current drawn from the battery.
Solution:
(a) Total resistance of the combination in parallel


(b) Potential of 20 V will be same across each resistor, so current

Question 5.
At room temperature (27.0°C) the resistance of a heating element is 100 O. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10^{3 0}C^-1.
Solution:

Question 6.
A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10^-7 m₂, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?
Solution:

Question 7.
A silver wire has a resistance of 2.1 Ω, at 27.5°C and a resistance of 2.7 Ω at 100°C. Determine the temperature coefficient of resistivity of silver.
Solution:

Question 8.
A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C ?Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10^-4 °C^-1.
Solution:
At room temperature 27 °C, the resistance of the heating element.

Question 9.
Determine the current in each branch of the network shown in figure.

Solution:
Marti Let us first distribute the current in different branches. Now, eΩ actions for different loops using Kirchhoff’s II^nd law,



Question 10.
(a) In a meter bridge as shown in figure, the balance point is found to be at 39.5 cm from the end A, when the resistor is of 12.5 Ω. Determine the resistance of x. Why are the connection between resistors in a Wheatstone or meter bridge made of thick copper strips?

(b) Determine the balance point of the bridge above if x and are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?
Solution:
(a)

The connections are made of thick copper strips so as to provide negligible resistance by connecting wires.
(b) If x and Y are interchanged the balance point will be at 60.5 cm from A.
(c) In balanced condition of the bridge, the cell and the galvanometer can be exchanged, the galvanometer will still show zero deflection.

Question 11.
A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Solution:

Now, terminal voltage of the battery during charging
V = E + lr = 8 + 7(0.5) = 11.5 V A series resistance is joined in the charging circuit to limit the excessive current so that charging is slow and permanent.

Question 12.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
Solution:
The potential gradient remains the same, as there is no change in the setting of standard circuit.

Question 13.
The number density of free electrons in a copper conductor estimated is ,8.5 × 10²⁸ m^-3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10^-6 m² and it is carrying a current of 3.0 A.
Solution:
We can first calculate drift velocity of the electrons from the given data I = Aneu_d

Question 14.
Theearth’s surface has a negative surface charge density of 10^-9  cm^-2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 10⁶ m).
Solution:
Due to negative charge on earth, an electric field is into the earth surface due to which the positive ions of atmosphere are constantly pumped in and an equivalent current of 1800 A is established across the globe. Let us first calculate total negative charge on earth

Question 15.
(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance of 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?

(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell ? Could the cell drive the starting motor of a car ?
Solution:
(a) Six cells are joined in series.

Equivalent emf is 2 × 6 = 12 V
Equivalent internal resistance is 0.015 × 6 = 0.09 Ω
Current drawn from supply

(b) Maximum current is drawn from a battery when external resistance is tested to be zero

To start a car, a current of the order of 100 A is needed, so the battery mentioned above can not drive the starting motor.

Question 16.
Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables.

Solution:
Two wires have same length, and resistance. As the specific resistances are unequal, the areas are different. For copper wire, R_cu = p_cu 

Thus, the aluminium wire for the same resistance is very light than copper and that is why aluminium wires are preferred for overhead power cables.

Question 17.
What conclusion can you draw from the following observations on a resistor made of alloy manganin?

Solution:
We can find resistance of the alloy manganin for all the readings as follows :

Here we can conclude that Ohm’s law is valid to a high accuracy and resistance of the alloy is nearly constant at all currents.

Question 18.
Answer the following questions:

1. A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?
2. Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.
3. A low voltage supply from which one needs high currents must have very low internal resistance. Why?
4. A high tension (HT) supply of, say, 6 kV must have a large internal resistance. Why?

Solution:

1. The current will be constant because it is given to be steady.
   
2. Ohm’s law is not a fundamental law in nature. It is not universally followed. Semiconductor diodes, transistors, their mistors, vacuum tubes etc. do not follow Ohm’s law.
3. If emf of supply battery is E and internal resistance r, then current through an external resistance R is given by  so, internal resistance r should be least to supply high currents.
4. In high tension supply, the internal resistance is made large. Because, if accidently the short circuiting take place, the excessive current produced should not cross the safety limits.

Question 19.
Choose the correct alternative:

1. Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
2. Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
3. The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature.
4. The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (1022 or 103).

Solution:

1. Alloys of metals usually have greater resistivity than that of their constituent metals.
2. Alloys usually have much lower temperature coefficients of resistance than pure metals.
3. The resistivity of the alloy manganin is nearly independent of increasing temperature.
4. The resistivity of a typical insulator (e.g. amber) is greater than that of a metal by factor of the order of 1022.

Question 20.
(a) Given n resistors each of resistance R, how will you combine them to get the

• maximum
• minimum effective resistance? What is the ratio of the maximum to minimum resistance?

(b) Given the resistances of 1 Ω, 2Ω, 3Ω, how will be combine them to get an equivalent resistance of

• (11/3) Ω
• (11/5) Ω,
• 6 Ω,
• (6/11) Ω?

(c) Determine the equivalent resistance of network shown in figure.

Solution:
(a)

• For maximum effective resistance, all the resistors should be joined in series. Rm_max = R + R + R+……..n or R_max = nR
• For minimum effective resistance, all the resistors should be joined in parallel.

(b) All possible combinations with resistances 1 Ω, 2 Ω and 3 Ω are


So, the combinations for the desired results can be selected.

Question 21.
Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown in figure. Each resistor has 1 Ω resistance.

As shown in diagram the loop of resistance shown by arrow is repeated at times, let us assume the equivalent resistance is x, so by adding one more loop the resistance will remain as x.

Question 22.
Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf E and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

1. What is the value of E?
2. What purpose does the high resistance of 600 kΩ have?
3. Is the balance point affected by this high resistance?
4. Is the balance point affected by the internal resistance of the driver cell?
5. Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?
6. Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

Solution:

1. For emf of 1.02 V, the balance point is 67.3 cm of wire. For unknown emf E, the balance point is 82.3 cm of wire.
   
2. A high resistance of 600 kΩ is joined in series with cell so as to prevent the galvanometer from excessive current, when the jockey is touched far from null point. Once the null point is closely known the 600 kΩ is short circuited so that exact position of null point can be achieved.
3. In presence of high resistance we will obtain null point for a longer length then a sharp position, it is because of very low current near null point.
4. Internal resistance of the cell do not affect the balance point, as no current is drawn at null point.
5. If the driver cell of standard circuit has an emf smaller than emf of the cell to be balanced then null point will not be achieved in the length of potentiometer wire and galvanometer will provide only one side deflection always.
6. In the given state, the circuit will be unsuitable, as for E of the order of mV the null point will be very close to end A and percentage error in the measurement of emf will be larger.
   To make suitable arrangement the potential drop in potentiometer wire is brought down to few mV, so that balance point is obtained at suitable length and this is done by introducing a high resistance in series in the standard calibration circuit.
   

Question 23.
Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance x is 68.5 cm. Determine the value of x. What might you do if you failed to find a balance point with the given cell of emfε?

Solution:
For comparison of resistances, R and x should be in series and same current should flow through them.

If we fail to find balance point with cell of e.m.f. e, in that case we should reduce the current I and for that e.m.f. E of cell responsible for current I should reduce, or a resistor can also be attached in series with R and x to reduce current.

Question 24.
Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Solution:
In the open circuit, the balance point is obtained for the emf of 1.5 V.

When the external circuit is connected, a current is drawn from the cell of 1.5 V in external resistance of 9.5 E1. Now the balance point is obtained for terminal potential

NCERT Exercises

Question 1.
Two charges 5 × 10^-88 C and -3 × 10^-8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. .
Solution:
(i) Let C be the point on the line joining the two charges, where electric potential is zero, then

So, electric potential is zero at distance of 10 cm from charge of 5 × 10^-8 C on line joining the two charges between them. If point C is not between the two charges, then


So, electric potential is also equal to zero a*a distance of 24 cm from charge of -3 × 10^-8 C and at a distance of (24 +16) = 40 cm from charge of 5 × 10^-8 C, on the side of charge of -3 × 10^-8 C.

Question 2.
A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the center of the hexagon.
Solution:

Question 3.
Two charges 2 μC and -2 μC are placed at points A and B 6 cm apart.

1. Identify an equipotential surface of the system.
2. What is the direction of the electric field at every point on this surface?

Solution:

1. Since it is an electric dipole, so a plane normal to AB and passing through its mid-point has zero potential everywhere.
2. Normal to the plane in the direction AB.

Question 4.
A spherical conductor of radius 12 cm has a charge of 1.6 × 10^-7 C distributed uniformly on its surface. What is the electric field

(a) inside the sphere
(b) just outside the sphere
(c) at a point 18 cm from the centre of the sphere?

Solution:

Question 5.
A parallel plate capacitor with air between the plates has a capacitance of 8 μF. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Solution:

Question 6.
Three capacitors each of capacitance 9 μF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitors if the combination is connected to a 120 V supply?

Solution:

Question 7.
Three capacitors of capacitances 2 μF, 3 μF and 4 μF are connected in parallel.

1. What is the total capacitance of the combination.
2. Determine the charge on each capacitor if the combination is connected to a 100 V supply.

Solution:

1. C = C₁ + C₂ + C₃ = 2 + 3 + 4 or C = 9 μF
2. Since the capacitors are in parallel, so potential difference across each of them is same i.e.

Question 8.
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^-3 m² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this, capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Solution:

Question 9.
Explain what would happen if in the capacitor given in above question, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates.

(a) While the voltage supply remained connected.
(b) After the supply was disconnected.

Solution:

Question 10.
A 12 μF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?
Solution:

Question 11.
A 600 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 μF capacitor. How much electrostatic energy is lost in the process?
Solution:

C, = 600 μF, V₁ = 200 V, C₂ = 600 μF, V₂ = 0

Question 12.
Acharge of 8 mC is located at theorigin. Calculate the work done in taking a small charge of – 2 × 10^-9 C from a point P(0,0,3 cm) to a point Q (0,4 cm, 0), via a point R (0,6 cm, 9 cm).
Solution:
As electric field is conservative field, so work done in moving a charge in electric field is independent of path chosen to move the charge in electric field and depends only on the electric potential difference between the two end points. So

Question 13.
A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.
Solution:
The length of diagonal of the cube of each side b is

E = 0, as electric field at centre due to a charge at any corner of cube is just equal and opposite to that of another charge at diagonally opposite corner of cube.

Question 14.
Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:

(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Solution:
(a) At mid point C of line joining two charges, electric potential is V_c = V_CA + V_CB

Question 15.
A spherical conducting shell of inner radius r, and outer radius r₂ has a charge Q.

1. A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
2. Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Solution:

1. Surface charge density on the inner surface of shell is
   
2. The electric flux linked with any closed surface S inside the conductor is zero as electric field inside conductor is zero.
   So, by Gauss’s theorem net charge enclosed by closed surface S is also zero i.e., 
   So, if there is no charge inside the cavity, then there cannot be any charge on the inner surface of the shell and hence electric field inside the cavity will be zero, even though the shell may not be spherical.
   

Question 16.
Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by  where n is a unit vector normal eo to the surface at a point and o is the surface charge density at that point. (The direction of n is from side 1 to side 2). Hence show that just
outside a conductor, the electric field is .
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
Solution:
Normal component of eDctric field intensity due to a thin infinite plane sheet of

(b) To show that the tangential component of electrostatic field is continuous from one side of a charged surface to another, we use the fact that work done by electrostatic field on a closed loop is zero.

Question 17.
A long charged cylinder of linear charge density k is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?
Solution:
In Figure, A is a long charged cylinder of linear charge density k, length l and radius a. A hollow co-axial conducting cylinder B of of length l and radius b surrounds A. The charge q = kl spreads uniformly on the outer surface of A. It induces – q charge on the cylinder B, which spreads on the inner surface of B. An electric field E is produced in the space between the two cylinders, which is directed radially outwards. Let us consider a co-axial cylindrical Gaussian surface of radius r. The electric flux through the cylindrical Gaussian surface is

The electric flux through the end faces of the cylindrical Gaussian surface is zero, as E is parallel to them. According to Gauss’s 

Question 18.
In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 A:
(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 A separation?
Solution:

Question 19.
If one of the two electrons of a H₂ molecule is removed, we get a hydrogen molecular ion . In the ground state of an , the two protons are separated by roughly 1.5 A, and the electron is roughly 1 A from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.
Solution:
It is a system of three point charges and the potential energy stored is this system of charges is

with zero potential energy at infinity.

Question 20.
TWO charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.
Solution:
As the two spheres are connected to each other by wire, so they have same electric potential i.e.,V_a = V_b

if b > a, then E_a > E_b
i.e. sphere with smaller radius produces more electric field on its surface. Hence, the charge density on the sharp and pointed ends of conductor is higher than on its flatter portions.

Question 21.
Two charges -q and +q are located at points (0,0, -a) and (0,0, a), respectively.

(a) What is the electrostatic potential at the points (0,0, z) and (x, y, 0)?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a »1.
(c) How much work is done in moving a small test charge from the point (5, 0, 0) to (-7, 0, 0) along the x-axis? Does the

answer change if the path of the test charge between the some points is not along the x-axis?
Solution:
(a) The two point charges form an electric dipole of moment p = q 2_a directed along + z-axis. Point A (0, 0, z) lies on the axis of electric dipole, so electric potential at point A is

Point B (x, y, 0) lies on the equatorial plane of electric dipole, so electric potential at point B is zero i.e. V_B = 0
(b) Electric potential at any point on the axis of electric dipole at distance r from its centre is

(c) As both the points C (5, 0, 0) and D (-7, 0, 0) lie on the perpendicular bisector of electric dipole, so electric potential at both the points is zero. Hence work done in moving the charge from C to D is
W_CD = q₀ [V_D – V_C] × O or W_CD = O
This work done will remain equal to zero even if the path of the test charge between the same points is changed, as electric field is conservative field and work done in moving a charge between the two points in electric field is independent of the path chosen to move the charge.

Question 22.
Figure shows a charge array known as an electric quadrupole. For a point on the axis of quadrupole, obtain the dependence of potential on r for r/a >> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

Solution:

Question 23.
An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1 kV. A large number of 1 μF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.
Solution:
Minimum number of capacitors that must be connected in series in a row are

Question 24.
What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm?
Solution:

Question 25.
Obtain the equivalent capacitance of the network in figure. For a 300 V supply, determine the charge and voltage across each capacitor.

Solution:

Question 26.
The plates of a parallel plate capacitor have an area of 90 cm² eacn and are separated by 2.5 mm.The capacitor is charged by connecting it to a 400 V supply
(a) How much electrostatic energy is stored by the capacitor?
(b) View this energy as stored in the electro-static field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.
Solution:

Question 27.
A 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
Solution:

Question 28.
Show that the force on each plate of a parallel plate capacitor has magnitude equal to  QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor .
Solution:
Magnitude of electric field between the
plates of charged capacitor is  However, magnitude of electric field of one plate or! the other plate of charged capacitor is  So, force on the one plate of charged capacitor due to the other is

The factor  is because the electric field of one 2 plate on the other plate of charged capacitor is  of the resultant electric field E between the 2 plates of charged capacitor.

Question 29.
A spherical capacitor consists of two concentric spherical conductors held in position by suitable insulating supports. Show that the capacitance of a spherical capacitor is given by  Where r₁ and r₂ are the radii of outer and inner spheres, respectively.
Solution:
Let + Q be the charge on outer spherical shell A of radius r, and + Q be the charge on inner spherical shell B of radius r2. Then electric potential on shell A is

This gives the capacitance of the spherical capacitor.

Question 30.
A spherical capacitor has an inner sphere of radius 12 cm and outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 μC. The space between the concentric spheres is filled with liquid of dielectric constant 32.

(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere.
(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
Solution:

Here C > C₀, because a single conductor A can be charged to a electric potential till it reaches the breakdown value of surroundings. But when another earthed metallic conductor B is ; brought near it, negative charge induced on it decreases the electric potential on A, hence more charge can not be stored on A.

Question 31.
Answer carefully:

1. Two large conducting spheres carrying charges Q₁ and Q₂ are brought close to each other, is the magnitude of electrostatic force between them exactly given by  r is the distance between their centres?
2. If coulomb’s law involved 1 /r³ dependence (instead of 1/r²), would Gauss’s law be still true?
3. A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
4. What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?
5. We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?
6. What meaning would you give to the capacitance of a single conductor?
7. Guess a possible reason why water has a much greater dielectric constant {- 80) then say, mica (= 6).

Solution:

1. No, because coulomb’s law holds good only for point charges. ‘
2. No, because in that case electric flux linked with the closed surface will also become dependent on V other than charge enclosed by it.
3. No, it will travel along the field line only if it is a straight line.
4. Zero, whatever may be the shape of orbit may be. It is because work done in moving a charge in closed path in electric field is zero, as electric field is a conservative field.
5. No, electric potential is continuous
   there. As E = 0, so  = 0 or V = constant.
6. It means that a single conductor is a capacitor whose other plate can be considered to be at infinity.
7. A water molecule is a polar molecule with non-zero electric dipole moment, however mica does not have polar molecules. So, dielectric constant of water is high.

Question 32.
A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 μC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e. bending of the field lines at the ends).
Solution:

Question 33.
A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 10⁷ V m^-1. For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 μF?
Solution:

Question 34.
Describe schematically the equipotential surfaces corresponding to

• A constant electric field in the z-direction.
• a field that uniformly increases in magnitude but remains in a constant (say, z) direction.
• a single positive charge at the origin, and
• a uniform grid consisting of long equally spaced parallel charged wires in a plane.

Solution:

• Planes parallel to x-y plane or normal to the electric field in z-direction.
• Planes parallel to x-y plane or normal to the electric field in z-direction, but the planes having different fixed potentials will become closer with increase in electric field intensity.
• Concentric spherical surfaces with their centres at origin.
• A time dependent changing shape nearer to grid, and at far off distances from the grid, it slowly becomes planar and parallel to the grid.

Question 35.
In a Van-de-Graff type generator, a spherical metal shell is to be a 15 × 106 V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 10⁷ V m¹. What is the minimum radius of the spherical shell required?
Solution:

Question 36.
A small sphere of radius r₁ and charge q₁ is enclosed by a spherical shell of radius r₂ and charge q₂. Show that if q: is positive, charge will necessary flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q² on the shell is.
Solution:
The potential on inner small sphere is VA

higher potential than outer conducting shell B, for any value of charge qv So, when inner sphere A is connected to outer shell B, then charge will flow from inner sphere A to outer shell B, until electric potentials on them is same i.e.
V_A – V_B = 0 or q₁ = 0 [As r₁ # r₂]

So, charge ql given to sphere A will flow on the shell B, no matter what the charge on the shell B is.

Question 37.
Answer the following:

1. The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 V m^-1. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage, so there is no field inside).
2. A man fixes outside house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1 m₂. Will he get an electric shock if he touches the metal sheet next morning?
3. The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?
4. What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning?

Solution:

1. Our body and the ground are at same electric potential. As we step out into the open, the original equipotential surfaces of open air change, keeping our head and the ground at the same potential.
2. Yes, it is because the aluminium sheet gets charged due to discharging current and raises to the extent depending on the capacitor formed by the sheet and the ground.
3. The atmosphere of earth gets continuously charged due to lightning, thunderstorms but simultaneously it gets discharged through normal weather zones. This keeps the system balanced.
4. Light, sound and heat energy. Light energy in lightning and heat and sound energy in the accompanying thunder.

NCERT Exercises

Question 1.
What is the force between two small charged spheres having charges of 2 × 10^-7 C and 3 × 10^-7 C placed 30 cm apart in air?
Solution:

Question 2.
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge -0.8 μC in air is 0.2 N.
(a)What is the distance between the two spheres?
(b)What is the force on the second sphere due to the first?
Solution:

Question 3.
Check that the ratio  is dimensionless. Look up a table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Solution:
The ratio of electrostatic force to the gravitational force between an electron and a proton separated by a distance r from each other is

This ratio signifies that electrostatic forces are 10³⁹ times stronger than gravitational forces.

Question 4.
(a) Explain the meaning of the statement ‘electric charge of a body is quantised’.
(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?
Solution:
(a) The net charge possessed by a body is an integral multiple of charge of an electron i.e. q = ± ne, where n = 0,1,2,3,… is the number of electrons lost gained by the body and e = 1.6 × 10^-19 C is charge of an electron. This is called law of quantization of charge.
(b) At macroscopic level, charges are enormously large as compared to the charge of an electron, e = 1.6 × 10^-19 C. Even a charge of 1 nC contains nearly 10¹³ electronic charges. So, at this large scale, charge can have a continuous value rather than discrete integral multiple of e, and hence, the quantization of electric charge can be ignored.

Question 5.
When a glass rod is rubbed with a silk cloth, charges appears on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
Solution:
When a glass rod is rubbed with a silk cloth, electrons from the glass rod are transferred to the piece of silk cloth. Due to this, the glass rod acquires positive (+) charge whereas the silk cloth acquires negative (-) charge. Before rubbing, both the glass rod and silk cloth are neutral and after rubbing the net charge on both of them is also equal to zero. Such similar phenomenon is observed with many other pairs of bodies. Thus, in an isolated system of bodies, charge is neither created nor destroyed, it is simply transferred from one body to the other. So, it is consistent with the law of conservation of charge.

Question 6.
Four point charges q_A = + 2 μC, q_B = -5μC, q_c = + 2 μC and q_D = -5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?
Solution:

Forces of repulsion on 1 μC charge at O due to 2 μC charge, at A and C are equal and opposite. Therefore, they cancel. Similarly, forces of attraction on 1 μC charge at O, due to -5 μC charges at B and at D are also equal and opposite. Therefore, these also cancel. Hence, the net force on the charge of 1 μC at O is zero.

Question 7.
(a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
(b)Explain why two field lines never cross each other at any point?
Solution:

A field line cannot have sudden breaks because the moving test charge never jumps from one position to the other.
(b)Two field lines never cross each other at any point/because if they do so, we will obtain two tangents pointing in two different directions of electric field at a point, which is not possible.

Question 8.
Two point charges q_A = 3 μC and q_B = -3 μC are located 20 cm apart in vacuum.
(a)What is the electric field at the midpoint O of the line AS joining the two charges?
(b)If a negative test charge of magnitude 1.5 × 10^-9 C is placed at this point, what is the force experienced by the test charge?
Solution:
(a) Electric fields at O due to the charges at A and B are

In direction opposite to that of E i.e. along OA.

Question 9.
A system has two charges q_A = 2.5 × 10^-7 C and q_B = -2.5 × 10^-7 C located at point A (0,0, -15 cm) and B (0,0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?
Solution:

Question 10.
An electric dipole with dipole moment 4 × 10^-9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10⁴ N C^-1. Calculate the magnitude of the torque acting on the dipole.
Solution:

Question 11.
A Polythene piece rubbed with wool is found to have a negative charge of 3 × 10^-7 C.
(a) Estimate the number of electrons transferred (from which to which?)
(b) Is there a transfer of mass from wool to polythene?
Solution:
(a) q = ne or n = 
= 1.875 × 10¹²
so, 1.875 × 10¹² electrons have transferred from wool to polythene, as polythene acquires negative charge.
(b) Yes, mass of 1.875 × 10¹² electrons i.e., m = nm_e = 1.875 × 10¹² × 9.1 × 10^-3 1 kg = 1.71 × 10^-18 kg has transferred from wool to polythene

Question 12.
(a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10^-7 C ? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
Solution:

Question 13.
Suppose the spheres A and B in previous question have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?
Solution:

Question 14.
Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Solution:
Particles 1 and 2 are negatively charged as they experience forces in direction opposite to that of electric field E, whereas particle 3 is positively charged as it experience force in the direction of electric field E.

Particle-3 has the highest charge to mass ratio, as it shows maximum deflection in the electric field.

Question 15.
Consider a uniform electric field .
(a)What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?
(b)What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?
Solution:

Question 16.
What is the net flux of the uniform electric field in previous question through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
Solution:
 As the net electric flux with closed surface like cube in uniform electric
field is equal to zero, because the number of lines entering the cube is the same as the number of lines leaving the cube.

Question 17.
Careful measurement of the electric field at the surface of a black box indicates that the net outward flux though the surface of the box is 8.0 × 10³ N m² C^-1.
(a)What is the net charge inside the box?
(b)If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?
Solution:

So, the net charge enclosed by that closed surface is zero, although it may have some charges inside it.

Question 18.
A point charge +10 μC is at a distance of 5 cm directly above the centre of a square of side 10 cm, as shown in figure. What is the magnitude of the electric flux though the square?
Solution:
Let us assume that the given square be one face of the cube of edge 10 cm. As charge of +10 μC is at a distance of 5 cm above the centre of a square, so it is enclosed by the cube. Hence by Gauss’s theorem, electric flux linked with the cube is

Question 19.
A point charge of 2.0 pC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
Solution:

Question 20.
A point charge causes an electric flux of -1.0 × 10³ Nm² C^-1 to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge?
Solution:
(a) On increasing the radius of the Gaussian surface, charge enclosed by it remains the same and hence the electric flux linked with Gaussian surface also remains the same.

Question 21.
A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 10³ N C^-1 and points radially inward. What is the net charge on the sphere?
Solution:

Question 22.
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC m^-2.
(a) Find the charge on the sphere.
(b) What is the total electric flux leaving the surface of the sphere?
Solution:

Question 23.
An infinite line charge produces a field of 9 × 10⁴NC^-1 at a distance of 2 cm. Calculate the linear charge density.
Solution:

Question 24.
Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitudes 17.0 × 10^-22 C m^-2. What is Electric field in
(a) the outer region of the first plate.
(b) the outer region of the second plate, and
(c) between the plates?
Solution:
As both the plates have same surface charge density σ, so

Question 25.
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ N C^-1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm^-3. Estimate the radius of the drop.
(g = 9.81 m s^-2; e= 1.6 × 10^-19 C)
Solution:
In equilibrium, force due to electric field on the drop balances the weight mg of drop

Question 26.
Which among the curves shown in Figure cannot possibly represent electrostatic field lines?
Solution:
(a) It is wrong, because electric field lines must be normal to the surface of conductor outside it.
(b) It is wrong because field lines cannot start or originate from negative charge, and also cannot end or submerge into positive charge.
(c) It is correct
(d) It is wrong because electric field lines never intersect each other.
(e) It is wrong because electric field lines cannot form closed loops.

Question 27.
In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 10⁵ N C^-1 per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10^-7 C m in the negative z-direction?
Solution:
As electric field increases in positive z-direction from A to B, So

directed in direction of FB i.e., from B to A or along -z direction.
As the two forces on charges of electric dipole are collinear and opposite, so net torque on it is equal to zero.

Question 28.
(a) A conductor A with a cavity is shown in figure (a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.
(b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q. (figure (b))
(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.
Solution:
(a) Let us consider a closed surface inside the conductor enclosing the cavity. As electric field inside the conductor is zero,

This shows that there cannot be any charge on the inner surface of conductor at the cavity, so any charge Q given to the conductor will appear only on its outer surface.
(b) Let us consider that the charge -q_I is induced on the inner surface of conductor A at the cavity, when conductor B of charge q is kept in its cavity. Due to this charge of Q + qt is induced on the outer surface of conductor. Let us construct a closed Gaussian surface S inside the conductor A enclosing its cavity. As, electric field inside conductor is zero, so

i.e., equal and opposite charge, -q is induced on inner surface of conductor A at the cavity,

when conductor B of charge, + q is kept is its cavity. Hence, the charge on outer surface of conductor is Q + q₁ = Q + q
(c) As we found that electric field inside the cavity of conductor is zero, even on charging the conductor, so the sensitive instrument can be shielded from the strong electrostatic fields in its environment, by covering it with a metallic cover.

Question 29.
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is  .where  is the unit vector in the outward normal direction, and o is the surface charge density near the hole.
Solution:
Inside a charged conductor, the electric field is zero.

But a uniformly charged flat surface provide

If we consider a small flat part on the surface of charged conductor, it certainly provides an 
Now if a hole is made in charged conductor, the fifeld due to small flat part is absent but the field due to rest of charged conductor is present i.e., equal to .

Question 30.
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density X without using Gauss’s law.
Solution:
Consider a point P, a unit away from the long charged wire. Electric field due to element dy,

vertical components cancel out and horizontal components are added due to symmetry

Question 31.
It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge + 2/3 e, and the ‘down’ quark (denoted by d) of charge (-1/3)e, together with electrons build up ordinary matter. [Quarks of other types have also been found which give rise to different unusual varieties of matter]. Suggest a possible quark composition of a proton and neutron.
Solution:

Question 32.
(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = o) of the configuration. Show that the equilibrium of the test charge is necessarily unstable, (b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed at certain distance apart.
Solution:
(a) Let us assume that a radial electric field is present with origin as centre. The E is 0 at origin and a small test charge ‘q0’ is placed at origin. Force on test charge at origin is zero.  But as the test charge is displaced a little, the electrostatic force will drift it further away so charge is in unstable equilibrium.

(b) A test charge placed at mid point of equal charges is in stable equilibrium, for lateral displacement. Initially at mid point position of test charge, two repulsion forces on test charge are equal and opposite.

But on displacing q₀ close to one of the charge, one of the forces become stronger and pushes the charge towards another.

F_A > F_B so q₀ moves towards B.

Question 33.
A particle of mass m and charge (-q) enters the region between the two charged plates initially moving along x-axis with speed vx.The length of plate is L and an uniform electric field £ is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL²/.
Solution:
Let the point at which the charged particle enters the electric field, be origin O (0, 0), then after travelling a horizontal displacement L, it gets deflected by displacement y in vertical direction as it comes out of electric field. So, co-ordinates of its initial position are x₁ = 0 and y₁ = 0 and final position on coming out of electric field are

and of initial velocity are u_x = v_x and u_y = 0 so, by 2^nd equation of motion in horizontal direction,

This gives the vertical deflection of the particle at the far edge of the plate.

Question 34.
Suppose that the particle in question 33 is an electron projected with velocity v_x = 2.0 × 10⁶ m s^-1. If E between the plates separated by 0.5 cm is 9.1 × 10² N C^-1 where will the electron strike the upper plate? (|e| = 1.6 × 10^-19, me = 9.1 × 10^-31 kg)
Solution:
If the electron is released just near the negatively charged plate, then y = 0.5 cm and hence