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08 Electrostatics

Multiple Choice Questions

1. Condenser is a device used to store……………………

(A) large charge at low potential

(B) large charge at high potential

(C) large charge at low potential

(D) less charge at high potential.

Ans. (C) large charge at low potential

2. Two condensers each of capacity

(A)
(B)
(C)
(D)

Ans. (C)

3. In the following figure, charge on capacitor is

(A)
(B)
(C)
(D)

Ans. (C)

4. Intensity of electric field at a point close and outside a charged conducting cylinder is proportional to…………………….

( is the distance of a point from the axis of cylinder)

(A)
(B)
(C)
(D)

Ans. (A)

5. Electric field intensity in free space at a distance ‘ ‘ outside the charged conducting sphere of radius ‘ ‘ in terms of surface charge density ‘ ‘ is

(A)
(B)
(C)
(D)

Ans. (A)

6. If the radius of a sphere is doubled without changing the charge on it, then electric flux originating from the sphere is…………………….
(A) double
(B) half
(C) same
(D) zero

Ans. (C) same

7. Electric intensity due to a charged sphere at a point outside the sphere decreases with…………………….

(A) increase in charge on sphere.

(B) increase in dielectric constant.

(C) decrease in the distance from the centre of sphere.

(D) Decrease in square of distance from the centre of sphere.

Ans. (B) increase in dielectric constant.

8. If the charge on the condenser of is doubled, then the energy stored in it becomes…………………….

(A) zero

(B) twice that of initial energy

(C) half the initial energy

(D) four times the initial energy

Ans. (D) four times the initial energy

9. Electric intensity outside a charged cylinder having the charge per unit length ‘ ‘ at a distance from its axis is…………………….

(A)
(B)
(C)
(D)

Ans. (C)

10. The electric field intensity outside the charged conducting sphere of radius ‘ ‘, placed in a medium of permittivity at a distance ‘ ‘ from the centre of the sphere in terms of surface charge density is

(A)
(B)
(C)
(D)

Ans. (A)

11. The angle at which maximum torque is exerted by the external uniform electric field on the electric dipole is …………………….

(A)
(B)
(C)
(D)

Ans. (D)

Theory Questions

8.2 Application of Gauss’ law

1. Obtain an expression for electric field intensity at a point outside a charged conducting sphere.

Ans:

i. Consider a sphere of radius with its centre at , charged to a uniform surface charge density placed in a dielectric medium of permittivity . The total charge on the sphere, .

ii. To find the electric field intensity at a point , at a distance from the centre of the charged sphere, imagine a concentric Gaussian sphere of radius passing through as shown in the figure below. Let ds be a small area around the point on the Gaussian surface.

Uniformly charged spherical shell or hollow sphere

iii. By Gauss’ theorem, the net flux through a closed Gaussian surface,

(for air/vacuum )

where is the total charge inside the closed surface.

iv. Due to symmetry and spheres being concentric, the electric field at each point on the Gaussian surface has the same magnitude and it is directed radially outward. Also, the angle between the direction of and the normal to the surface of the sphere (ds) is zero i.e.,

Flux through the area

Total electric flux through the Gaussian surface

v. From equations (1) and (2),

This is the expression for the electric intensity at the point , outside the charged sphere.

2. Obtain an expression for electric field intensity at a point outside uniformly charged thin plane sheet.

Ans:

i. Consider a uniformly charged infinite plane sheet with uniform surface charge density (charge per unit area) kept in a medium of permittivity .

ii. To find the electric field due to a charged infinite plane sheet at at a distance from sheet, imagine a Gaussian surface around in the form of a cylinder having cross sectional area and length with its axis perpendicular to the plane sheet. The plane sheet passes through the middle of the length of the cylinder sach that the ends of the cylinder (called end caps and ) are equidistant (at a distance ) from the plane sheet as shown in the figure below.

iii. By Gauss’ theorem, the net flux through a closed surface

for air/vacuum

where is the total charge inside the closed surface.

iv. By symmetry electric field is perpendicular to plane sheet and directed outwards, having same magnitude at a given distance on either sides of the sheet. The electric field is at right angles to the end caps and away from the plane. Its magnitude is the same at and . The flux passing through the curved surface is zero as the electric field is tangential to this surface.

The total flux through the closed Gaussian surface is given by,

v. From equations (1) and (2),

This is the required expression.

vi. Direction of electric field is outward if sheet is positively charged and inward if it is negatively charged.

3. State Gauss’ theorem in electrostatics. State the expression for electric field intensity at a point outside an infinitely long charged conducing cylinder.

Ans:

i. Gauss’ law: The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by .

where is the total charge within the surface.

ii. Expression for electric field intensity at a point outside an infinitely long charged conducing cylinder:

4. State the formula for electric field intensity at a point outside an infinitely long charged cylindrical conductor.

Ans: Refer Subtopic 8.2: Q. No. 3 (ii)

8.8 Dielectrics and Electric Polarisation

1. With the help of neat diagram, explain how non-polar dielectric material is polarised in external electric field of increasing intensity. Define polarisation in dielectrics.

Ans:

i. In the presence of an external electric field , the centres of the positive charge in each molecule of a non-polar dielectric is pulled in the direction of , while the centres of the negative charges are displaced in the opposite direction. Thus, the two centres are separated and the molecule gets distorted.

ii. The displacement of the charges stops when the force exerted on them by the external field is balanced by the restoring force between the charges in the molecule.

iii. Each molecule becomes a tiny dipole having a dipole moment.

iv. The induced dipole moments of different molecules add up giving a net dipole moment to the dielectric in the presence of the external field.

v. Polarisation: Certain substances when are placed in an external field, their positive and negative charges get displaced in opposite directions and the molecules develop a net dipole moment. This is called polarization of the material.

2. What are polar dielectrics and non-polar dielectrics?

Ans:

i. Polar dielectric:

a. A dielectric molecule in which the centre of mass of positive charges (protons) does not coincide with the centre of mass of negative charges (electrons), because of the asymmetric shape of the molecules is called polar dielectric.

b. They have permanent dipole moments of the order of . They act as tiny electric dipoles, as the charges are separated by a small distance.

c. Examples: , water, alcohol,

ii. Non-polar dielectric:

A dielectric in which the centre of mass of the positive charges coincides with the centre of mass of the negative charges is called a nonpolar dielectric.

b. These have symmetrical shapes and have zero dipole moment in the normal state.

c. Examples: , benzene, methane

8.9 Capacitors and Capacitance, Combination of Capacitors in Series and Parallel

1. Define capacitance of a capacitor and its S.I. unit.

Ans:

i. The ability of a conductor to store the electric charge is called capacitance of conductor.

ii. SI unit of capacity of conductor is farad (F).

Thus, the capacity of a conductor is said to be 1 farad if the potential difference across it rises by 1 volt, when charge is given to it.

2. Obtain an expression for equivalent capacitance of two capacitors and connected in series.

Ans:

i. Diagram:

ii. Explanation:

a. Capacitors are said to be connected in series if they are connected one after the other in the form of a chain.

b. Let capacitors of capacitances be connected in series as shown in the figure.

c. Let be the corresponding potential differences in the capacitors.

d. Suppose a potential difference ‘ ‘ is applied across the combination. The left plate of capacitor has a charge . An equal but opposite charge is induced on the right plate of this capacitor. This charge induces a charge on the left plate of .

e. Thus, each capacitor receives a magnitude of charge Q. Hence, when the capacitors are connected in series, same current flows through them and all have the same charge induced on the plate.

Thus, potential difference induced across capacitors is given by,

f. Total potential difference ‘ ‘ across the combination is given by,

….[From equation (1)]

g. If these capacitors are replaced by a single capacitor of capacity , such that effective capacity remains same then

From equation (2) and (3),

8.10 Capacitance of a Parallel Plate Capacitor Without and With Dielectric Medium Between the Plates

1. Draw neat, labelled diagram of a parallel plate capacitor with a dielectric slab between the plates.

Ans:

8.12 Energy Stored in a Capacitor

1. Obtain an expression for energy of a charged capacitor and express it in different forms.

Ans:

i. Consider a capacitor of capacitance being charged by a DC source of V volts as shown in figure below.

ii. During the process of charging, let be the charge on the capacitor and be the potential difference between the plates. Hence
iii. A small amount of work is done if a small charge is further transferred between the plates.

iv. Total work done in transferring the charge

v. This work done is stored as electrical potential energy of the capacitor. This work done can be expressed in different forms as follows.

8.13 Van de Graaff Generator

1. Describe the construction of van-de-Graaff generator.
Ans: Construction:

i. It consists of a large hollow metallic sphere D mounted on two insulating columns (not shown in the diagram given below).

ii. There are two pulleys and is mounted at the centre of sphere while is mounted near the bottom. A long narrow belt made of insulating material such as rubber or silk passes over the pulleys. The belt BB is driven by an electric motor (not shown in the diagram), connected to the lower pulley .

iii. The spray brush A, consisting of a large number of pointed wires, is connected to the positive terminal of a high voltage DC power supply. From this brush positive charge can be sprayed on the belt which can be collected by another similar brush connected to .

iv. is an evacuated accelerating tube having an electrode I at its upper end, connected to the dome-shaped conductor.

v. To prevent the leakage of charge from the dome, the pulley and belt arrangement, the dome and a part of the evacuated tube are enclosed inside a large steel vessel , filled with nitrogen at high pressure. A small quantity of Freon gas is mixed with nitrogen to ensure better insulation between the vessel and its contents.

vi. A metal plate held opposite to the brush on the other side of the belt is connected to the vessel , which is earthed.

2. With the help of a neat diagram, describe the construction and working of Van de Graaff generator.

Ans: Construction: Refer Subtopic 8.13: Q. No. 1

Working:

i. The electric motor connected to the pulley is switched on, to set the conveyor belt into motion. The DC supply is then switched on.

ii. From the pointed ends of the spray brush A, positive charge is continuously sprayed on the belt .

iii. The belt carries this charge in the upward direction, which is collected by the collector brush C and sent to the dome shaped conductor.

iv. As the dome is hollow, the charge is distributed over the outer surface of the dome. Its potential rises to a very high value due to the continuous accumulation of charges on it. The potential of the electrode I also rises to this high value.

v. The positive ions such as protons or deuterons from a small vessel (not shown in the figure) containing ionised hydrogen or deuterium are then introduced in the upper part of the evacuated accelerator tube.

vi. These ions, repelled by the electrode I, are accelerated in the downward direction due to the very high fall of potential along the tube, these ions acquire very high energy.

vii. These high energy charged particles are then directed so as to strike a desired target.

Numericals

8.2 Application of Gauss’ law

1. A conductor of any shape, having area placed in air is uniformly charged with a charge . Determine the electric intensity at a point just outside its surface. S. I. units

Solution:

Given:

To find: i. Electric field intensity (E)

ii. Mechanical force per unit area (f)

Formulae: i. ii.

Calculation:

i. Using formula (i) and (iii),

ii. From formula (ii)

Ans: i. Electric intensity is

ii. Mechanical force per unit area is 141.2 N/m

8.5 Equipotential Surfaces

1. A small particle carrying a negative charge of is suspended in equilibrium between two horizontal metal plates apart, having a potential difference of 4000 volts across them. Find the mass of the particle.

Solution:

Given: ,

To find: Mass of the particle (m)

Formula:

Calculation: From formula

As the charged particle remain suspended, in equilibrium,

Ans: Mass of the particle is .

8.6 Electrostatic Potential Energy of Two Point Charges and of a Dipole in an Electrostatic Field

1. An electric dipole consists of two opposite charges each of magnitude , separated by . The dipole is placed in an external electric field of .

Calculate the:

i. maximum torque experienced by the dipole and

ii. work done by the external field to turn the dipole through .

Solution:

Given:

To find: i. Maximum torque

ii. Work done in rotating dipole

Formulae: i.

ii.

iii.

Calculation: From formula (i)

For maximum torque,

As and

Ans: i. The maximum torque exerted by field on dipole is .

ii. Work done in rotating dipole is .

8.9 Capacitors and Capacitance, Combination of Capacitors in Series and Parallel

1. A parallel plate air condenser has an area and separation between the two plates is . Find its capacity.

[Mar 08]

Solution:

Given:

To find: Capacitance (C)

Formulae:

Calculation: From formula,

Ans: Capacity of given capacitor is .

2. A parallel plate capacitor has circular plates, each of diameter separated by a distance of . The potential difference between the plates is maintained at . Calculate its capacitance and charge. What is the intensity of electric filed between the plates of the capacitor? (Given : )

Solution:

Given:

To find:

i. Capacitance (C)

ii. Charge (Q)

iii. Electric field intensity

Formulae: i.

ii.

iii.

Calculation: From formula (i),

From formula (ii),

From formula (iii),

Ans: i. Capacitance is .

ii. Charge is .

iii. Electric field intensity is .

3. A network of four capacitors of each is connected to a supply. Determine the charge on each capacitor.

Solution:

Given:

,

To find: The charge on each capacitor , and )

Formulae: i.

ii.

iii.

Calculation: The equivalent network for the given circuit is as follows:

Using formula (i),

equivalent capacity of three capacitors in series,

is connected in parallel to the combination of and .

Using formula (ii),

Equivalent capacitance,

Let potential across be given by respectively.

Also, Hence, .

Also charges on respectively are same.

The charge on capacitor is and is given by,

Ans: The charge on is and that on is .

4. Six capacitors of capacities and are connected as shown in the network given below.

Find:

i. The value of if the network is balanced, and

ii. The resultant capacitance between and .

Solution:

Given:

To find:

A balanced network of capacitors with

,

i. The value of

ii. Resultant capacitance between and

Formulae:
i.

ii.

(For balance condition)

iii.

Calculation: For branch DC of the given network Using formula (i),

Now using formula (ii),

As the network is balanced, the circuit can be reduced as follows:

For path ,

For part ADC,

Using formula (iii)

Hence,

Ans: i. The value of is .

ii. The resultant capacitance between A and is .

5. Three capacitors of capacities and are connected in a series and potential difference of 120 volt is maintained across the combination. Calculate the charge on capacitor of capacity . 

Solution:

Given:

To find: Charge on the capacitor

Formulae: i.

Calculation: Using formula (i),

In series combination,

Using formula (ii),

Ans: The charge on the capacitor is .

6. A parallel plate capacitor filled with air has an area of and plate separation of a . Calculate its capacitance.

Solution:

Given:

To find: Capacitance of capacitor (C)

Formulae:

Calculation: From formula,

Ans: Capacitance is .

8.10 Capacitance of a Parallel Plate Capacitor Without and With Dielectric Medium Between the Plates

1. A parallel plate air condenser has a capacity of . What will be the new capacity if:

i. the distance between the two plates is doubled?

ii. a marble slab of dielectric constant 8 is introduced between the two plates?

Solution:

Given:

To find: i. The new capacity if distance between two plates is doubled

ii. The new capacity if a marble slab is introduced between two plates

Formula:

Calculation:

i. From formula,

To find: Electrostatic energy (U)

ii.

[ Distance between plates remains same]

Ans: i. The new capacity will be if distance between two plates is doubled.

ii. The new capacity will be if a marble slab is introduced between two plates.

2. Capacity of a parallel capacitor with dielectric constant 5 is . Calculate the capacity of the same capacitor when dielectric material is removed.

[Mar 20]

Solution:

Value of dielectric constant,

Capacitance without dielectric,

since,

Ans: The capacity of capacitor without dielectric material is .

8.12 Energy Stored in a Capacitor

1. Energy stored in a charged capacitor of is . Find the charge on its plate.

Solution:

Given:

To find: Electrostatic energy (U)

Formula:

Calculation: From formula,

Ans: The charge on its plate is

2. A condenser of capacity is charged to a potential of . Calculate the energy stored in the condenser.

Solution:

Given:

Formula:

Calculation: From formula,

Ans: The electrostatic energy stored in the capacitor is .

3. A cube of marble having each side is kept in an electric field of intensity . Determine the energy contained in the cube of dielectric constant 8.

Solution:

[Given: ] [Mar 17]

Given:

To find: ‘ Energy contained in the cube (U)

Formula:

Calculation: Volume of marble,

Energy density,

From formula,

Ans: The energy contained in the cube is .

4. A capacitor is connected with battery. What would be the electrostatic energy stored? [July 18]

Solution:

Given:

To find: Electrostatics energy stored (U)

Formula:

Calculation: Using Formula,

Ans: The electrostatic energy stored in capacitor is .