Electrostatic Potential and Capacitance

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Learning Objectives

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Learning Objectives:
- Understand the concept of electrostatic potential and its relation to electric fields.
- Calculate the electric potential due to point charges and charge distributions.
- Analyze the behavior of capacitors in series and parallel configurations.
- Apply the principles of electrostatics to solve problems involving capacitors and electric fields.
- Explore the effects of dielectric materials on capacitance and electric fields.
- Investigate the concept of electrostatic shielding and its applications.

Revision Notes & Summary

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Electrostatic Potential and Capacitance

Introduction

The notion of potential energy was introduced in previous chapters, where work done against a force results in stored potential energy.
Coulomb force between stationary charges is a conservative force, similar to gravitational force.

Key Concepts

Electrostatic Shielding: A cavity inside a conductor is shielded from outside electrical influences, but charges inside the cavity do not shield the exterior.

Exercises

Electric Potential Zero: Two charges 5 X 10⁻⁸ C and -3 X 10⁻⁸ C are located 16 cm apart. Find points where electric potential is zero.
Hexagon Charge Potential: A regular hexagon with side 10 cm has a charge of 5 µC at each vertex. Calculate the potential at the center.
Equipotential Surface: Two charges 2 µC and -2 µC are placed 6 cm apart. Identify an equipotential surface and the direction of the electric field on this surface.
Spherical Conductor: A spherical conductor of radius 12 cm has a charge of 1.6 X 10⁻⁷C. Calculate the electric field inside, just outside, and at a point 18 cm from the center.
Capacitance Calculation: A parallel plate capacitor with air has a capacitance of 8 pF. Calculate capacitance if the distance is halved and filled with a dielectric constant of 6.
Series Capacitors: Three capacitors of 9 pF each are connected in series. Find total capacitance and potential difference across each capacitor connected to a 120 V supply.
Parallel Capacitors: Three capacitors of 2 pF, 3 pF, and 4 pF are connected in parallel. Find total capacitance and charge on each capacitor connected to a 100 V supply.
Capacitance of Parallel Plate: For a parallel plate capacitor with area 6 X 10⁻³ m² and distance 3 mm, calculate capacitance and charge when connected to a 100 V supply.

Important Formulas

Capacitance: C = Q/V, where Q is charge and V is potential difference.
Capacitance in Series: $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \cdots$
Capacitance in Parallel: C = C₁ + C₂ + C₃ + ...
Energy Stored in Capacitor: U = $\frac{1}{2} C V^2$
Electric Field in Dielectric: E = E₀/K, where K is the dielectric constant.

Points to Ponder

Electrostatics deals with forces between charges at rest, maintained by unspecified opposing forces.
A capacitor confines electric field lines, resulting in a small potential difference despite strong fields.
Electric field is zero inside a charged spherical shell, while potential is continuous across the surface.
The torque on a dipole in an electric field causes oscillation, aligning with the field if damped.
Potential due to a charge at its own location is infinite.

Exam Tips & Common Mistakes

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Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Electrostatic Shielding: Students often think that charges inside a conductor's cavity shield the exterior from external electric fields. Remember, the exterior is not shielded from fields created by inside charges.
Confusion in Calculating Electric Potential: When asked to find points where electric potential is zero between two charges, students may forget to consider points outside the two charges.
Capacitance Miscalculations: When capacitors are connected in series, students sometimes incorrectly add their capacitances instead of using the formula for series combinations.

Tips for Success

Visualize Charge Distributions: Draw diagrams to help understand how charges are distributed on conductors and the resulting electric fields.
Practice with Exercises: Regularly solve exercises like calculating electric potential at various points and determining capacitance in different configurations to solidify understanding.
Review Key Formulas: Familiarize yourself with essential formulas such as capacitance for series and parallel combinations, and potential energy equations.
Understand the Concept of Equipotential Surfaces: Recognize that the electric field is always perpendicular to equipotential surfaces, which can help in visualizing electric field directions.

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Practice Test – MCQs, True/False

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Multiple Choice Questions

8 cm from the positive charge

12 cm from the negative charge

4 cm from the positive charge

10 cm from the negative charge

Correct Answer: A

Solution:

The electric potential is zero at a point where the potentials due to both charges cancel each other. Using the formula for potential,

V = k \frac{q}{r}

, we set the potentials equal and solve for the distance.

The charge on the capacitor increases.

The charge on the capacitor decreases.

The charge on the capacitor remains the same.

The charge on the capacitor becomes zero.

Correct Answer: A

Solution:

When a dielectric is inserted while the voltage supply remains connected, the capacitance increases, and since the voltage is constant, the charge on the capacitor increases according to the relation

Q = CV

25 V

50 V

75 V

100 V

Correct Answer: B

Solution:

Initially, the charge on the first capacitor is

Q = CV = 900 \times 10^{-12} \times 100 = 9 \times 10^{-8}

C. When connected to the uncharged capacitor, the total charge is shared equally, so

V' = \frac{Q}{C_{total}} = \frac{9 \times 10^{-8}}{1800 \times 10^{-12}} = 50

1.5 x 10^-9 J

3.0 x 10^-9 J

1.2 x 10^-9 J

2.4 x 10^-9 J

Correct Answer: A

Solution:

The energy stored in a capacitor is given by the formula

E = \frac{1}{2} C V^2

. Substituting the given values,

E = \frac{1}{2} \times 12 \times 10^{-12} \times (50)^2 = 1.5 \times 10^{-9} \text{ J}

4.5 µJ

9.0 µJ

0.45 µJ

45 µJ

Correct Answer: A

Solution:

The energy stored in a capacitor is given by

U = \frac{1}{2} C V^2

. Here,

C = 900 \times 10^{-12}

F and

V = 100

V. Thus,

U = \frac{1}{2} \times 900 \times 10^{-12} \times (100)^2 = 4.5 \times 10^{-6}

J or 4.5 µJ.

-6 x 10^{-3} J

-3 x 10^{-3} J

3 x 10^{-3} J

6 x 10^{-3} J

Correct Answer: A

Solution:

The potential energy

U

is given by

U = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1 q_2}{r}

. Substituting the given values,

U = \frac{9 \times 10^9 \cdot 2 \times 10^{-6} \cdot (-2 \times 10^{-6})}{0.06} = -6 \times 10^{-3} J

9 pF

8 pF

7 pF

6 pF

Correct Answer: A

Solution:

For capacitors in parallel, the total capacitance

C_{\text{total}} = C_1 + C_2 + C_3 = 2 + 3 + 4 = 9 \text{ pF}

4.5 x 10^-6 J

9.0 x 10^-6 J

2.25 x 10^-6 J

1.125 x 10^-6 J

Correct Answer: A

Solution:

The energy stored in each capacitor initially is

\frac{1}{2} C V^2 = \frac{1}{2} \times 900 \times 10^{-12} \times (100)^2 = 4.5 \times 10^{-6} \text{ J}

. Since they are connected in parallel and the total capacitance is doubled, the energy remains the same for the system.

100 V

200 V

50 V

150 V

Correct Answer: A

Solution:

When the charged capacitor is connected to an uncharged capacitor of the same capacitance, the charge is equally distributed between the two capacitors. Initially, the charge on the first capacitor is

Q = C \times V = 600 \times 10^{-12} \times 200 = 1.2 \times 10^{-7}

C. The total charge remains the same, but the effective capacitance becomes

C_{eq} = 600 + 600 = 1200

pF. The final voltage

V_f = \frac{Q}{C_{eq}} = \frac{1.2 \times 10^{-7}}{1200 \times 10^{-12}} = 100

qV(r)

\frac{1}{2}qV(r)

q^2V(r)

\frac{q}{V(r)}

Correct Answer: A

Solution:

The potential energy of a charge

q

in an external potential

V(r)

is given by

qV(r)

1.2 x 10^-5 J

2.4 x 10^-5 J

3.6 x 10^-5 J

4.8 x 10^-5 J

Correct Answer: A

Solution:

Initially, the energy stored is

E_1 = \frac{1}{2} C V^2 = \frac{1}{2} \times 600 \times 10^{-12} \times (200)^2 = 1.2 \times 10^{-5} \text{ J}

. After connecting to another capacitor, the total capacitance becomes 1200 pF and the energy stored is

E_2 = \frac{1}{2} \times 1200 \times 10^{-12} \times (100)^2 = 6 \times 10^{-6} \text{ J}

. Energy lost is

E_1 - E_2 = 1.2 \times 10^{-5} - 6 \times 10^{-6} = 6 \times 10^{-6} \text{ J}

27 pF

3 pF

9 pF

1 pF

Correct Answer: B

Solution:

For capacitors in series, the total capacitance

C

is given by

\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

. Substituting the given values,

\frac{1}{C} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3}

, so

C = 3 \text{ pF}

96 pF

48 pF

24 pF

12 pF

Correct Answer: A

Solution:

The capacitance of a capacitor is given by

C = \varepsilon_0 \frac{A}{d}

. If the distance is halved and a dielectric is introduced, the new capacitance becomes

C' = K \times C \times 2 = 6 \times 8 \times 2 = 96 \text{ pF}

1.25 J

2.5 J

3.75 J

5.0 J

Correct Answer: A

Solution:

The energy stored in a capacitor is given by

U = \frac{1}{2} C V^2

. Substituting

C = 10 \times 10^{-6} \, \text{F}

and

V = 500 \, \text{V}

, we get

U = \frac{1}{2} \times 10 \times 10^{-6} \times 500^2 = 1.25

7.5 µF

10 µF

12.5 µF

15 µF

Correct Answer: A

Solution:

The series combination of three capacitors, each 10 µF, gives an equivalent capacitance of

C' = \frac{10}{3} \text{ µF}

. This is in parallel with the fourth capacitor, so the total capacitance is

C = C' + 10 = \frac{10}{3} + 10 = 7.5 \text{ µF}

9 pF

2.5 pF

3.5 pF

4.5 pF

Correct Answer: A

Solution:

For capacitors in parallel, the total capacitance is the sum of the individual capacitances:

C_{\text{total}} = C_1 + C_2 + C_3 = 2 + 3 + 4 = 9

pF.

11.1 J/m³

0.11 J/m³

0.22 J/m³

0.055 J/m³

Correct Answer: D

Solution:

The energy density

u

of an electric field is given by

u = \frac{1}{2} \varepsilon_0 E^2

. With

\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m}

and

E = 5 \text{ N/C}

u = \frac{1}{2} \times 8.85 \times 10^{-12} \times (5)^2 = 0.055 \text{ J/m}^3

Charge remains constant

Charge increases

Charge decreases

Charge becomes zero

Correct Answer: B

Solution:

When a dielectric is inserted while the voltage supply remains connected, the capacitance increases by a factor equal to the dielectric constant. Since

Q = CV

, the charge increases as the capacitance increases while the voltage remains constant.

24 pF

96 pF

48 pF

16 pF

Correct Answer: B

Solution:

The original capacitance

C_0 = 8

pF. When the distance is halved, the capacitance doubles, so

C_1 = 2 \times 8 = 16

pF. With a dielectric constant

K = 6

, the new capacitance is

C = K \times C_1 = 6 \times 16 = 96

pF.

C_0

\frac{(K+1)}{2}C_0

KC_0

\frac{K}{2}C_0

Correct Answer: B

Solution:

When a dielectric slab fills half the space between the plates, the system can be considered as two capacitors in series: one with dielectric constant

K

and the other with air. The effective capacitance is given by

C = \frac{(K+1)}{2}C_0

It remains the same.

It doubles.

It increases by a factor of 6.

It decreases by a factor of 6.

Correct Answer: C

Solution:

The capacitance of a capacitor with a dielectric is given by

C = KC_0

, where

K

is the dielectric constant. Thus, the capacitance increases by a factor of 6.

C_0

K \cdot C_0

C_0/K

K^2 \cdot C_0

Correct Answer: B

Solution:

When a dielectric is inserted, the capacitance increases by a factor of the dielectric constant. Therefore, the new capacitance is

K \cdot C_0

-pE

pE

\frac{1}{2}pE

2pE

Correct Answer: A

Solution:

The potential energy of a dipole in a uniform electric field is given by

U = -p \cdot E

, where

p

is the dipole moment and

E

is the electric field.

\frac{1}{2} \varepsilon_0 E^2

\varepsilon_0 E^2

\frac{1}{2} E^2

E^2

Correct Answer: A

Solution:

The energy density of an electric field is given by

u = \frac{1}{2} \varepsilon_0 E^2

6 \times 10^{-6} \text{ J}

3 \times 10^{-6} \text{ J}

1.5 \times 10^{-6} \text{ J}

0 \text{ J}

Correct Answer: B

Solution:

Initially, the energy stored in the charged capacitor is

U_1 = \frac{1}{2} C V^2 = \frac{1}{2} \times 600 \times 10^{-12} \times (200)^2 = 1.2 \times 10^{-5} \text{ J}

. When connected to another uncharged capacitor, the total capacitance becomes

C_{total} = 600 + 600 = 1200

pF. The final energy is

U_2 = \frac{1}{2} C_{total} V'^2

, where

V'

is the new potential difference. Since charge is conserved,

Q = C V = C_{total} V'

, giving

V' = 100

V. Thus,

U_2 = \frac{1}{2} \times 1200 \times 10^{-12} \times (100)^2 = 6 \times 10^{-6} \text{ J}

. The energy lost is

U_1 - U_2 = 1.2 \times 10^{-5} - 6 \times 10^{-6} = 6 \times 10^{-6} \text{ J}

0 \text{ N/C}

5.33 \times 10^{4} \text{ N/C}

8.00 \times 10^{4} \text{ N/C}

1.07 \times 10^{5} \text{ N/C}

Correct Answer: B

Solution:

The electric field outside a charged spherical conductor is given by

E = \frac{kQ}{r^2}

, where

k = 8.99 \times 10^9 \text{ Nm}^2\text{/C}^2

Q = 1.6 \times 10^{-7} \text{ C}

, and

r = 0.18 \text{ m}

. Thus,

E = \frac{8.99 \times 10^9 \times 1.6 \times 10^{-7}}{(0.18)^2} = 5.33 \times 10^{4} \text{ N/C}

Increases by a factor of

K

Decreases by a factor of

K

Remains the same

Increases by a factor of

K+1

Correct Answer: A

Solution:

When a dielectric slab is inserted into a capacitor, the capacitance increases by a factor of the dielectric constant

K

, i.e.,

C = KC_0

where

C_0

is the original capacitance.

2700 V

5400 V

10800 V

1350 V

Correct Answer: B

Solution:

The potential at the center due to one charge is

V = \frac{k \cdot q}{r}

, where

k = 9 \times 10^9

Nm²/C²,

q = 5 \times 10^{-6}

C, and

r = 0.1

m. Hence,

V = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{0.1} = 450000

V. Since there are 6 charges, the total potential is

6 \times 450000 = 2700000

V or 2700 V.

1.2 x 10^4 N/C

2.4 x 10^4 N/C

3.6 x 10^4 N/C

4.8 x 10^4 N/C

Correct Answer: A

Solution:

The electric field just outside a charged spherical conductor is given by

E = \frac{kQ}{r^2}

. Substituting the given values,

E = \frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.12)^2} = 1.2 \times 10^4 N/C

4.5 x 10^{-6} J

9.0 x 10^{-6} J

2.25 x 10^{-6} J

1.8 x 10^{-6} J

Correct Answer: A

Solution:

The energy stored by the capacitor is calculated using the formula

E = \frac{1}{2} CV^2

. Substituting the given values,

E = \frac{1}{2} \times 900 \times 10^{-12} \times (100)^2 = 4.5 \times 10^{-6} J

15 \times 10^{-9} \text{ J}

1.5 \times 10^{-9} \text{ J}

3.0 \times 10^{-9} \text{ J}

6.0 \times 10^{-9} \text{ J}

Correct Answer: B

Solution:

The electrostatic energy stored in a capacitor is given by the formula

U = \frac{1}{2} C V^2

. Substituting the given values,

U = \frac{1}{2} \times 12 \times 10^{-12} \text{ F} \times (50 \text{ V})^2 = 1.5 \times 10^{-9} \text{ J}

1.77 \times 10^{-12} \text{ F}

1.77 \times 10^{-11} \text{ F}

1.77 \times 10^{-10} \text{ F}

1.77 \times 10^{-9} \text{ F}

Correct Answer: A

Solution:

The capacitance of a parallel plate capacitor is given by

C = \frac{\varepsilon_0 A}{d}

, where

\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m}

A = 6 \times 10^{-3} \text{ m}^2

, and

d = 3 \times 10^{-3} \text{ m}

. Thus,

C = \frac{8.85 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}} = 1.77 \times 10^{-12} \text{ F}

4 cm from the

5 \times 10^{-8}

C charge

8 cm from the

5 \times 10^{-8}

C charge

12 cm from the

5 \times 10^{-8}

C charge

16 cm from the

5 \times 10^{-8}

C charge

Correct Answer: A

Solution:

The potential is zero at a point where the potentials due to both charges cancel each other. Let the distance from the

5 \times 10^{-8}

C charge be

x

. Then,

\frac{k \times 5 \times 10^{-8}}{x} = \frac{k \times 3 \times 10^{-8}}{16-x}

. Solving,

x = 4

cm.

2.7 x 10^6 V

5.4 x 10^6 V

1.8 x 10^6 V

3.6 x 10^6 V

Correct Answer: B

Solution:

The potential at the center due to one charge is

V = \frac{k \cdot q}{r}

, where

k = 9 \times 10^9 \text{ Nm}^2/\text{C}^2

q = 5 \times 10^{-6} C

, and

r = 10 \times 10^{-2} m

. Thus,

V = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{0.1} = 4.5 \times 10^6 V

. Since there are 6 charges, the total potential is

6 \times 4.5 \times 10^6 = 2.7 \times 10^7 V

-pE \cos \theta

pE \sin \theta

pE \cos \theta

-pE \sin \theta

Correct Answer: A

Solution:

The potential energy

U

of a dipole in a uniform electric field is given by

U = -\mathbf{p} \cdot \mathbf{E} = -pE \cos \theta

, where

p = 2qa

is the dipole moment and

\theta

is the angle between the dipole moment and the electric field.

17.7 pF

1.77 pF

0.177 pF

177 pF

Correct Answer: A

Solution:

The capacitance

C

is given by

C = \varepsilon_0 \frac{A}{d}

. Substituting the values,

C = 8.85 \times 10^{-12} \times \frac{6 \times 10^{-3}}{3 \times 10^{-3}} = 17.7 \times 10^{-12} \text{ F} = 17.7 \text{ pF}

3 pF

9 pF

27 pF

1 pF

Correct Answer: A

Solution:

For capacitors in series, the total capacitance

C_{total}

is given by

\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

. Here,

C_1 = C_2 = C_3 = 9

pF, so

C_{total} = \frac{9}{3} = 3

pF.

200 pC, 300 pC, 400 pC

100 pC, 150 pC, 200 pC

200 pC, 250 pC, 300 pC

300 pC, 400 pC, 500 pC

Correct Answer: A

Solution:

In a parallel combination, the charge on each capacitor is given by

Q = CV

. For the 2 pF capacitor,

Q_1 = 2 \times 10^{-12} \times 100 = 200 \text{ pC}

. For the 3 pF capacitor,

Q_2 = 3 \times 10^{-12} \times 100 = 300 \text{ pC}

. For the 4 pF capacitor,

Q_3 = 4 \times 10^{-12} \times 100 = 400 \text{ pC}

. Thus, the charges are 200 pC, 300 pC, and 400 pC respectively.

True or False

Correct Answer: True

Solution:

Inserting a dielectric between the plates of a capacitor reduces the electric field, which in turn reduces the potential difference across the plates. This is due to the polarization of the dielectric material, which creates an opposing electric field.

Correct Answer: True

Solution:

In electrostatics, the electric field inside a cavity within a conductor is zero if there are no charges inside the cavity, due to the shielding effect of the conductor.

Correct Answer: True

Solution:

Inserting a dielectric material between the plates of a capacitor increases its capacitance because the dielectric reduces the effective electric field, allowing the capacitor to store more charge at the same voltage.

Correct Answer: False

Solution:

The potential energy stored in a system of two point charges is inversely proportional to the distance between them, given by

U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r}

Correct Answer: True

Solution:

The energy stored in a capacitor is indeed given by the formula

U = \frac{1}{2}CV^2

, where

C

is the capacitance and

V

is the potential difference across the capacitor.

Correct Answer: True

Solution:

Inside a conductor, the electric field is zero. This also applies to a cavity within the conductor, provided there are no charges inside the cavity.

Correct Answer: True

Solution:

The capacitance of a capacitor is a geometric property determined by the shapes, sizes, and relative positions of the conductors.

Correct Answer: True

Solution:

The energy density

u

of an electric field is given by

u = \frac{1}{2} \varepsilon_0 E^2

, where

\varepsilon_0

is the permittivity of free space and

E

is the electric field strength.

Correct Answer: False

Solution:

For capacitors in series, the reciprocal of the total capacitance

C

is the sum of the reciprocals of the individual capacitances:

\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots

Correct Answer: True

Solution:

The electric field inside a conductor is zero because charges reside on the surface, and the field inside cancels out.

Correct Answer: True

Solution:

The energy stored in a capacitor is indeed in the form of an electric field between its plates. The energy density of the electric field is given by

u = \frac{1}{2} \varepsilon_0 E^2

, where

E

is the electric field strength.

Correct Answer: True

Solution:

The energy density of an electric field is calculated as

\frac{1}{2} \varepsilon_0 E^2

, where

\varepsilon_0

is the permittivity of free space and

E

is the electric field.

Correct Answer: True

Solution:

The energy stored in a capacitor depends only on the final charge configuration and not on the process by which the charges were transferred. This is because electrostatic forces are conservative.

Correct Answer: True

Solution:

Inside a conductor, the electric field is zero because charges reside on the surface, and any internal field would cause charges to move until equilibrium is reached.

Correct Answer: False

Solution:

The electric field inside a dielectric is weaker than the field without the dielectric because the dielectric reduces the effective field between the plates.

Correct Answer: True

Solution:

The capacitance of a parallel plate capacitor is given by

C = \varepsilon_0 \frac{A}{d}

. When the distance

d

is halved and a dielectric with constant

K = 6

is introduced, the capacitance becomes

C = K \varepsilon_0 \frac{A}{d/2} = 12 \varepsilon_0 \frac{A}{d}

, which is an increase.

Correct Answer: False

Solution:

Inserting a dielectric material into a capacitor increases its capacitance. This is because the dielectric reduces the electric field within the capacitor, allowing it to store more charge for the same voltage.

Correct Answer: True

Solution:

The energy stored in a capacitor is given by the formula

\frac{1}{2} C V^2

. Substituting

C = 12 \times 10^{-12} F

and

V = 50 V

, we get

\frac{1}{2} \times 12 \times 10^{-12} \times 50^2 = 1.5 \times 10^{-9} J

Correct Answer: True

Solution:

The electric potential due to a dipole decreases with distance as

1/r^3

, which is faster than the

1/r^2

decrease typical of a single charge.

Correct Answer: True

Solution:

In electrostatic equilibrium, the electric field inside a conductor is zero, and any excess charge resides on the surface, as charges repel each other and move to the surface.

Correct Answer: True

Solution:

When a dielectric slab is inserted into a capacitor while the voltage supply remains connected, the dielectric reduces the electric field, which increases the capacitance.

Correct Answer: True

Solution:

The potential energy of two point charges in electrostatics is calculated using the formula

U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{12}}

, which is derived from Coulomb's law.

Correct Answer: True

Solution:

In a series combination of capacitors, the total capacitance is given by the reciprocal of the sum of the reciprocals of the individual capacitances, which is always less than any single capacitance in the series.

Correct Answer: True

Solution:

The potential energy of a dipole in a uniform electric field is given by the equation

U = -\mathbf{p} \cdot \mathbf{E}

, where

\mathbf{p}

is the dipole moment and

\mathbf{E}

is the electric field.

Correct Answer: True

Solution:

The potential energy

U

of a dipole in a uniform electric field

E

is given by

U = -p \cdot E

, where

p

is the dipole moment vector.

Correct Answer: True

Solution:

For capacitors in series, the total capacitance is given by

\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots

. This results in a total capacitance that is less than any individual capacitance.

Correct Answer: True

Solution:

The capacitance of a parallel plate capacitor is given by

C = \varepsilon_0 \frac{A}{d}

, where

A

is the area of the plates and

d

is the separation between them. If the distance is reduced by half and a dielectric with constant 6 is inserted, the capacitance increases by a factor of 12.

Correct Answer: False

Solution:

When the charged capacitor is connected to an uncharged capacitor, the total charge is redistributed, and the energy is shared between the two capacitors. This results in a loss of energy due to redistribution, not an increase.

Correct Answer: False

Solution:

The potential energy of a dipole in a uniform electric field is given by the equation

U = -\mathbf{p} \cdot \mathbf{E}

, where

\mathbf{p}

is the dipole moment and

\mathbf{E}

is the electric field.

Correct Answer: True

Solution:

The potential energy of a dipole in a uniform electric field is calculated as

-p \cdot E

, where

p

is the dipole moment and

E

is the electric field.

Correct Answer: True

Solution:

The energy stored in a capacitor is given by

U = \frac{1}{2} CV^2

, where

C

is the capacitance and

V

is the voltage across the capacitor, showing that it is directly proportional to the square of the voltage.

Correct Answer: True

Solution:

In electrostatics, the electric field inside a conductor is zero, and any excess charge resides on the surface of the conductor.

Correct Answer: False

Solution:

The effective capacitance of capacitors in series is always less than the smallest individual capacitance. This is because the reciprocal of the total capacitance is the sum of the reciprocals of the individual capacitances.

Correct Answer: False

Solution:

In a series combination, the reciprocal of the total capacitance is the sum of the reciprocals of the individual capacitances:

\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots

Correct Answer: True

Solution:

The energy density of an electric field is given by

u = \frac{1}{2} \varepsilon_0 E^2

and is a general result that holds for any configuration of charges.

Correct Answer: True

Solution:

Inserting a dielectric material between the plates of a capacitor increases its capacitance because the dielectric reduces the effective electric field between the plates, allowing more charge to be stored for the same potential difference.

Correct Answer: True

Solution:

Inserting a dielectric material like mica between the plates of a capacitor while the voltage supply remains connected increases the capacitance. The dielectric constant of the material affects the capacitance by a factor equal to the dielectric constant.

Correct Answer: True

Solution:

The electric field inside a conductor is zero because the charges reside on the surface, and any excess charge distributes itself on the surface to maintain electrostatic equilibrium.

Correct Answer: True

Solution:

The potential energy

U

of two point charges is given by

U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r}

, where

r

is the distance between the charges, indicating an inverse relationship.

Correct Answer: True

Solution:

When a dielectric is inserted between the plates of a capacitor, it induces a net dipole moment in the dielectric, which reduces the electric field and thus the potential difference across the plates.

Correct Answer: True

Solution:

The potential energy

U

of two charges

q_1

and

q_2

separated by a distance

r_{12}

is given by

U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{12}}

. This shows that the potential energy is inversely proportional to the distance

r_{12}

Correct Answer: True

Solution:

The energy density formula

\frac{1}{2}\varepsilon_0 E^2

is a general result that applies to any electric field configuration, not just parallel plate capacitors.

Correct Answer: True

Solution:

Initially, the energy stored in the charged capacitor is

\frac{1}{2} C V^2

. When connected to an uncharged capacitor, the total energy is redistributed, and due to the conservation of charge, some energy is lost as heat or radiation.

Correct Answer: True

Solution:

The potential energy

U

of two point charges

q_1

and

q_2

separated by a distance

r_{12}

is given by

U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{12}}

, which shows that it is inversely proportional to the distance

r_{12}

Correct Answer: True

Solution:

This formula represents the potential energy stored in a system of two point charges separated by a distance

r_{12}

Correct Answer: False

Solution:

For capacitors connected in series, the reciprocal of the total capacitance is the sum of the reciprocals of the individual capacitances, not the direct sum of the capacitances.

Correct Answer: True

Solution:

In a series combination, the reciprocal of the total capacitance is the sum of the reciprocals of the individual capacitances, resulting in a total capacitance that is less than any individual capacitance in the series.

Correct Answer: True

Solution:

When a dielectric slab with a thickness less than the plate separation is inserted, the electric field in the dielectric is reduced by a factor of the dielectric constant, which increases the capacitance.

Correct Answer: True

Solution:

The energy stored in a capacitor is calculated using the formula

\frac{1}{2} CV^2

, representing the work done to charge the capacitor.

Correct Answer: True

Solution:

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances:

C = C_1 + C_2 + C_3 + \, \ldots

Correct Answer: True

Solution:

For capacitors in series, the total capacitance is calculated using the formula

\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots

Correct Answer: True

Solution:

The potential energy

U

of a charge

q

in an external potential

V(r)

is given by the equation

U = qV(r)

. This relationship is fundamental in electrostatics.

Correct Answer: True

Solution:

The potential energy

U

of a charge

q

in an external potential

V(r)

is indeed given by

U = qV(r)

. This is a standard expression for potential energy in an electric field.

Correct Answer: True

Solution:

The electrostatic potential due to a dipole at a point is indeed the sum of the potentials due to the two charges of the dipole, as potential follows the superposition principle.

Correct Answer: True

Solution:

The formula for the electrostatic energy stored in a capacitor is

\frac{1}{2} CV^2

. This represents the energy stored due to the electric field between the plates.

Correct Answer: True

Solution:

The energy density of an electric field is given by

u = \frac{1}{2}\varepsilon_0 E^2

, which is a general result and holds true for any configuration of charges.

Correct Answer: False

Solution:

In a series combination of capacitors, the reciprocal of the total capacitance is the sum of the reciprocals of the individual capacitances:

\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots

Correct Answer: True

Solution:

This expression represents the potential energy of a charge

q

placed in an external electric potential

V(r)

Correct Answer: True

Solution:

When a dielectric material is inserted between the plates of a capacitor while the voltage supply remains connected, the capacitance increases due to the dielectric reducing the effective electric field between the plates.

Correct Answer: True

Solution:

In electrostatic equilibrium, the electric field inside a conductor is zero. This is because charges redistribute themselves on the surface of the conductor to cancel any internal electric fields.

Correct Answer: True

Solution:

In electrostatic equilibrium, the electric field inside a conductor is zero because charges reside on the surface, and any internal field would cause movement of charges.

Electrostatic Potential and Capacitance

Learning Objectives

Revision Notes & Summary

Electrostatic Potential and Capacitance

Introduction

Key Concepts

Exercises

Important Formulas

Points to Ponder

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Success

AI Learning Assistant

Practice Test – MCQs, True/False

Multiple Choice Questions

Q1.Two charges 5 x 10^{-8} C and -3 x 10^{-8} C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero?

Correct Answer: A

Q2.A parallel plate capacitor with air between the plates has a capacitance of 8 pF. If a 3 mm thick mica sheet with a dielectric constant of 6 is inserted between the plates while the voltage supply remains connected, what happens to the charge on the capacitor?

Correct Answer: A

Q3.A 900 pF capacitor is charged by a 100 V battery. It is then disconnected and connected to another 900 pF capacitor. What is the final potential difference across each capacitor?

Correct Answer: B

Q4.A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?

Correct Answer: A

Q5.What is the energy stored in a 900 pF capacitor charged by a 100 V battery?

Correct Answer: A

Q6.What is the potential energy of two charges 2 µC and -2 µC placed 6 cm apart?

Correct Answer: A

Q7.Three capacitors of capacitances 2 pF, 3 pF, and 4 pF are connected in parallel. What is the total capacitance of the combination?

Correct Answer: A

Q8.Consider two capacitors, each of 900 pF, initially charged by a 100 V battery. After disconnecting the battery, they are connected in parallel. What is the total electrostatic energy stored in the system?

Correct Answer: A

Q9.A 600 pF capacitor is charged by a 200 V supply and then disconnected. It is connected to another uncharged 600 pF capacitor. What is the final potential difference across each capacitor?

Correct Answer: A

Q10.What is the potential energy of a charge qqq in an external potential V(r)V(r)V(r)?

Correct Answer: A

Q11.A 600 pF capacitor is charged by a 200 V supply and then disconnected. It is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Correct Answer: A

Q12.What is the total capacitance of three capacitors each of capacitance 9 pF connected in series?

Correct Answer: B

Q13.A parallel plate capacitor with air between the plates has a capacitance of 8 pF. What will be the capacitance if the distance between the plates is reduced by half and the space is filled with a dielectric of constant 6?

Correct Answer: A

Q14.What is the energy stored in a capacitor with capacitance C=10 μFC = 10 \, \mu\text{F}C=10μF connected to a 500 V500 \, \text{V}500V supply?

Correct Answer: A

Q15.A network of four 10 µF capacitors is connected to a 500 V supply. If three of the capacitors are connected in series and the fourth is connected in parallel to this series combination, what is the equivalent capacitance of the network?

Correct Answer: A

Q16.Three capacitors of capacitances 2 pF, 3 pF, and 4 pF are connected in parallel. What is the total capacitance of the combination?

Correct Answer: A

Q17.What is the energy density of an electric field with a field strength of 5 N/C in a vacuum?

Correct Answer: D

Q18.If a mica sheet with a dielectric constant of 6 is inserted into a capacitor while the voltage supply remains connected, what happens to the charge on the capacitor?

Correct Answer: B

Q19.A parallel plate capacitor with air between the plates has a capacitance of 8 pF. If the distance between the plates is reduced by half and the space is filled with a dielectric of constant 6, what is the new capacitance?

Correct Answer: B

Q20.In a parallel plate capacitor, a dielectric slab of dielectric constant KKK is inserted such that it fills half the space between the plates. If the original capacitance was C0C_0C0​, what is the new capacitance?

Correct Answer: B

Q21.What happens to the capacitance of a parallel plate capacitor if a 3 mm thick mica sheet with a dielectric constant of 6 is inserted between the plates, while the voltage supply remains connected?

Correct Answer: C

Q22.A slab of material with dielectric constant KKK is inserted between the plates of a parallel-plate capacitor, reducing the electric field to E=E0/KE = E_0/KE=E0​/K. What is the new capacitance if the original capacitance was C0C_0C0​?

Correct Answer: B

Q23.What is the potential energy of a dipole moment ppp in a uniform electric field EEE?

Correct Answer: A

Q24.What is the energy density of an electric field with permittivity ε0\varepsilon_0ε0​ and electric field strength EEE?

Correct Answer: A

Q25.A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Correct Answer: B

Q26.Consider a spherical conductor of radius 12 cm with a charge of 1.6×10−71.6 \times 10^{-7}1.6×10−7 C distributed uniformly on its surface. What is the electric field at a point 18 cm from the center of the sphere?

Correct Answer: B

Q27.If a dielectric slab of constant KKK is inserted into a parallel-plate capacitor, how does the capacitance change?

Correct Answer: A

Q28.In a hexagonal arrangement of charges, each vertex has a charge of 5 µC. If the side of the hexagon is 10 cm, what is the potential at the center of the hexagon?

Correct Answer: B

Q29.A spherical conductor of radius 12 cm has a charge of 1.6 x 10^{-7} C distributed uniformly on its surface. What is the electric field just outside the sphere?

Correct Answer: A

Q30.A 900 pF capacitor is charged by a 100 V battery. How much electrostatic energy is stored by the capacitor?

Correct Answer: A

Q31.A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?

Correct Answer: B

Q32.Consider a parallel plate capacitor with air between the plates, each plate having an area of 6×10−3 m26 \times 10^{-3} \text{ m}^26×10−3 m2 and the distance between the plates being 3 mm. Calculate the capacitance of the capacitor.

Correct Answer: A