Chapter 17: Electrostatic Potential and Capacitance

Introduction

Concepts

• Test charge \( q_0 > 0 \) moved slowly from point R to P against repulsive force by fixed charge \( q > 0 \) at origin.
• Work done by external force \( W_{RP} = \int_R^P \vec{F}_{ext} \cdot d\vec{r} = - \int_R^P \vec{F}_E \cdot d\vec{r} \).
• Work done against electrostatic repulsion stores as potential energy.
• Potential energy difference: \( \Delta U = U_P - U_R = W_{RP} \).
• Work done by electrostatic field depends only on initial and final points.
• Only potential energy difference is physically significant.

Solved Examples

• Ex 1: Calculate work done by external force moving \( q_0 \) from R to P against \( q \).
  • Use \( W_{RP} = - \int_R^P \vec{F}_E \cdot d\vec{r} \).
  • Calculate electric force \( \vec{F}_E = k \frac{q q_0}{r^2} \).
  • Integrate over path from R to P.

Practice Set

• Q1: Work done by external force moving positive charge against repulsive force is
  • A) Positive and stored as potential energy
  • B) Negative and lost as heat
  • C) Zero
  • D) Equal to kinetic energy gained
• Q2: Potential energy difference between two points depends on
  • A) Path taken
  • B) Initial and final points only
  • C) Speed of charge
  • D) Magnitude of test charge only
• Q3: Work done by electrostatic force moving charge from R to P is
  • A) Equal to work done by external force
  • B) Negative of work done by external force
  • C) Zero
  • D) Independent of path

Answer Key

• Q1: A - Work done by external force is positive and stored as potential energy.
• Q2: B - Depends only on initial and final points, path independent.
• Q3: B - Work by electrostatic force is negative of external force work.

Memory Aids

• Work done by external force = potential energy stored.
• Potential energy difference depends only on start and end points.
• Electrostatic force work = negative of external force work.

Electrostatic Potential

Concepts

• Electrostatic potential \( V \) at a point = work done by external force to bring unit positive charge from infinity to that point.
• Reference potential at infinity \( V_\infty = 0 \).
• Potential at point \( P \): \( V_P = \frac{W_P}{q_0} \).
• Potential difference between points A and B:
  • \( V_B - V_A = \frac{W_{AB}}{q_0} = - \int_A^B \vec{E} \cdot d\vec{l} \)
• Electric field \( \vec{E} \) points away from positive charge.

Solved Examples

• Ex 1: Find potential at distance \( r \) from point charge \( +q \).
  • Use \( V = k \frac{q}{r} \).
• Ex 2: Calculate potential difference between points A and B.
  • Use \( V_B - V_A = - \int_A^B \vec{E} \cdot d\vec{l} \).
  • Evaluate integral along path.

Practice Set

• Q1: Electrostatic potential at infinity is
  • A) Zero
  • B) Infinite
  • C) Equal to potential at point P
  • D) Negative
• Q2: Potential difference between two points is work done per unit charge in moving charge
  • A) From infinity to point
  • B) From A to B against electrostatic force
  • C) Along any path
  • D) Only in vacuum
• Q3: Potential at distance 0.5 m from charge 2 \( \mu C \) (\( k=9\times10^9 \)) is
  • A) 36 kV
  • B) 18 kV
  • C) 72 kV
  • D) 9 kV

Answer Key

• Q1: A - Potential at infinity is defined as zero.
• Q2: B - Work done per unit charge moving from A to B against electrostatic force.
• Q3: A - \( V = kq/r = 9\times10^9 \times 2\times10^{-6} / 0.5 = 36,000 V = 36 kV \).

Memory Aids

• Potential \( V = \frac{W}{q_0} \), work done per unit charge.
• Potential at infinity = 0 (reference).
• Potential difference = negative line integral of electric field.

Potential due to a Point Charge

Concept and Formula

• Electric potential \( V \) at distance \( r \) from point charge \( q \) is work done per unit charge in bringing a test charge from infinity to that point.
• Work done \( W = -q_0 \int_{\infty}^{r} E \, dx = -q_0 \int_{\infty}^{r} \frac{1}{4\pi\epsilon_0} \frac{q}{x^2} dx \)
• Evaluating integral: \( W = \frac{1}{4\pi\epsilon_0} \frac{q q_0}{r} \)
• Electric potential: \( V = \frac{W}{q_0} = \frac{1}{4\pi\epsilon_0} \frac{q}{r} \)
• Corresponding electric field: \( E(r) = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} = \frac{V(r)}{r} \)

Solved Examples

• Ex 1: Calculate potential at 0.5 m from a point charge of 2 \( \mu C \).
  • Use \( V = \frac{1}{4\pi\epsilon_0} \frac{q}{r} \)
  • \( q = 2 \times 10^{-6} C, r = 0.5 m \)
  • \( V = 9 \times 10^9 \times \frac{2 \times 10^{-6}}{0.5} = 36 \times 10^3 V = 36 kV \)
• Ex 2: Find electric field at 0.2 m from a charge of 5 \( \mu C \).
  • \( E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{(0.2)^2} = 1.125 \times 10^6 V/m \)

Practice Set

• Q1: Electric potential at 1 m from a point charge of 1 \( \mu C \) is:
  • A) 9 V
  • B) 9000 V
  • C) 900 V
  • D) 90 V
• Q2: Electric field at 0.1 m from a charge of 3 \( \mu C \) is:
  • A) \( 2.7 \times 10^6 \) V/m
  • B) \( 3 \times 10^5 \) V/m
  • C) \( 2.7 \times 10^5 \) V/m
  • D) \( 3 \times 10^6 \) V/m
• Q3: Potential at 0.25 m from a charge \( 4 \times 10^{-6} C \) is:
  • A) 144 kV
  • B) 72 kV
  • C) 36 kV
  • D) 18 kV

Answer Key

• Q1: B) 9000 V
  Calculation: \( V = 9 \times 10^9 \times \frac{1 \times 10^{-6}}{1} = 9000 V \)
• Q2: A) \( 2.7 \times 10^6 \) V/m
  Calculation: \( E = 9 \times 10^9 \times \frac{3 \times 10^{-6}}{(0.1)^2} = 2.7 \times 10^6 V/m \)
• Q3: A) 144 kV
  Calculation: \( V = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{0.25} = 144000 V = 144 kV \)

Memory Aids

• Potential \( V \) due to point charge: \( V \propto \frac{1}{r} \)
• Electric field \( E \) due to point charge: \( E \propto \frac{1}{r^2} \)
• Relation: \( E = \frac{V}{r} \)
• Constant \( \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \) Nm²/C²

Potential due to an Electric Dipole

Axial Position Potential

• Dipole: two charges +q and -q separated by distance 2a.
• Point C lies along the axis beyond positive charge.
• Potential at C: \( V_C = \frac{1}{4\pi\epsilon_0} \left( \frac{-q}{r+a} + \frac{q}{r - a} \right) \)
• Simplified: \( V_C = \frac{1}{4\pi\epsilon_0} \frac{2aq}{r^2 - a^2} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2 - a^2} \), where \( p = 2aq \) is dipole moment.

Equatorial Position Potential

• Point C lies on the perpendicular bisector of dipole at distance r.
• Potential at C: \( V_C = \frac{1}{4\pi\epsilon_0} \left( \frac{-q}{\sqrt{r^2 + a^2}} + \frac{q}{\sqrt{r^2 + a^2}} \right) = 0 \)
• Potential is zero at equatorial line of dipole.

Potential at Any Point Due to Short Dipole

• Let OP = r, distances from charges to point P are \( r - a\cos\theta \) and \( r + a\cos\theta \).
• Potential at P: \( V_P = \frac{1}{4\pi\epsilon_0} \left( \frac{-q}{r + a\cos\theta} + \frac{q}{r - a\cos\theta} \right) \)
• Simplified: \( V_P = \frac{1}{4\pi\epsilon_0} \frac{2aq\cos\theta}{r^2 - a^2 \cos^2\theta} = \frac{1}{4\pi\epsilon_0} \frac{p\cos\theta}{r^2 - a^2 \cos^2\theta} \)
• For \( r \gg a \), \( V_P \approx \frac{1}{4\pi\epsilon_0} \frac{p\cos\theta}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \hat{r}}{r^2} \)

Solved Examples

• Ex 1: Calculate potential at axial point 10 cm from dipole center with \( p = 4 \times 10^{-29} \) C·m. Use \( \epsilon_0 = 8.85 \times 10^{-12} \) F/m.
  • Use \( V = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2} \)
  • \( V = \frac{9 \times 10^9 \times 4 \times 10^{-29}}{(0.1)^2} = 3.6 \times 10^{-8} \) V
• Ex 2: Find potential at equatorial point 5 cm from dipole center with \( q = 1 \times 10^{-6} \) C and \( 2a = 2 \times 10^{-2} \) m.
  • Potential at equatorial line is zero.
• Ex 3: Calculate potential at point P at 15 cm from dipole center making \( 60^\circ \) with dipole axis, \( p = 3 \times 10^{-29} \) C·m.
  • Use \( V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2} \)
  • \( V = 9 \times 10^9 \times \frac{3 \times 10^{-29} \times 0.5}{(0.15)^2} = 6 \times 10^{-9} \) V

Practice Set

• Q1: Potential at axial point 20 cm from dipole with dipole moment \( 6 \times 10^{-29} \) C·m is:
  • A) \( 1.35 \times 10^{-8} \) V
  • B) \( 1.35 \times 10^{-7} \) V
  • C) \( 3.0 \times 10^{-8} \) V
  • D) Zero
• Q2: Potential at equatorial point of dipole is:
  • A) Maximum
  • B) Zero
  • C) Depends on distance
  • D) Negative
• Q3: Potential at point P at distance 10 cm from dipole center, angle \( 45^\circ \), dipole moment \( 4 \times 10^{-29} \) C·m is:
  • A) \( 1.27 \times 10^{-8} \) V
  • B) \( 2.54 \times 10^{-8} \) V
  • C) Zero
  • D) \( 9.0 \times 10^{-9} \) V

Answer Key

• Q1: A; \( V = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2} = 9 \times 10^9 \times \frac{6 \times 10^{-29}}{0.2^2} = 1.35 \times 10^{-8} \) V.
• Q2: B; Potential at equatorial line is always zero due to equal and opposite contributions.
• Q3: A; \( V = 9 \times 10^9 \times \frac{4 \times 10^{-29} \times \cos 45^\circ}{0.1^2} = 1.27 \times 10^{-8} \) V.

Memory Aids

• Axial potential \( \propto \frac{p}{r^2} \), maximum along dipole axis.
• Equatorial potential = 0, due to symmetry.
• General potential: \( V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2} \) for \( r \gg a \).
• Dipole moment \( p = q \times 2a \), vector from -q to +q.

Potential due to a system of charges

Concept and Formula

• Electric potential (V) is the potential energy per unit charge at a point (unit: Volt, V = J/C).
• Potential is a scalar quantity; it has magnitude only, no direction.
• Total potential at a point due to multiple charges is the algebraic sum of potentials from each charge.
• Formula: \( V = k \sum_{i=1}^n \frac{Q_i}{r_i} \), where \(k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \ \text{Nm}^2/\text{C}^2\).
• \(Q_i\) = charge magnitude, \(r_i\) = distance from charge to point.

Electric Potential at a Point from Multiple Charges

• Identify charges and their signs (+ or -).
• Measure distances \(r_i\) from each charge to the point of interest.
• Calculate individual potentials \(V_i = k \frac{Q_i}{r_i}\).
• Sum all potentials algebraically: \(V_p = V_1 + V_2 + V_3 + \ldots + V_n\).

General Formula for Multiple Charges

• Electric potential at point \(P\) due to \(n\) charges: \( V_p = \frac{1}{4\pi\epsilon_0} \sum_{i=1}^n \frac{q_i}{r_i} \).
• \(\epsilon_0\) is permittivity of free space.
• Use summation notation for compactness and clarity.

Solved Examples

Ex 1

• Given two charges \(q_1 = +2 \mu C\) at 0.5 m and \(q_2 = -3 \mu C\) at 0.3 m from point P.
• Calculate total potential at P: \(V_p = k \left( \frac{q_1}{r_1} + \frac{q_2}{r_2} \right)\).
• Substitute values: \(V_p = 9 \times 10^9 \left( \frac{2 \times 10^{-6}}{0.5} + \frac{-3 \times 10^{-6}}{0.3} \right)\).
• Calculate and sum to find \(V_p\).

Practice Set

Q1

Two point charges +4 \(\mu C\) and +6 \(\mu C\) are placed 0.2 m and 0.3 m from point P respectively. What is the total electric potential at P? (Use \(k=9 \times 10^9\))

• A) 3.0 \times 10^5 V
• B) 4.2 \times 10^5 V
• C) 5.4 \times 10^5 V
• D) 6.0 \times 10^5 V

Q2

Three charges +2 \(\mu C\), -3 \(\mu C\), and +1 \(\mu C\) are at distances 0.1 m, 0.2 m, and 0.3 m from point P. Calculate the net potential at P.

• A) 1.35 \times 10^5 V
• B) -1.35 \times 10^5 V
• C) 0 V
• D) 2.7 \times 10^5 V

Q3

Four charges +1 \(\mu C\), +2 \(\mu C\), -1 \(\mu C\), and -2 \(\mu C\) are placed at distances 0.1 m, 0.2 m, 0.3 m, and 0.4 m from point P respectively. Find the total potential at P.

• A) 0 V
• B) 4.5 \times 10^4 V
• C) -4.5 \times 10^4 V
• D) 9.0 \times 10^4 V

Answer Key

• Q1: C) 5.4 \times 10^5 V
  Calculation: \(V_p = 9 \times 10^9 \left( \frac{4 \times 10^{-6}}{0.2} + \frac{6 \times 10^{-6}}{0.3} \right) = 5.4 \times 10^5 V\).
• Q2: A) 1.35 \times 10^5 V
  Sum potentials: \(9 \times 10^9 \left( \frac{2 \times 10^{-6}}{0.1} - \frac{3 \times 10^{-6}}{0.2} + \frac{1 \times 10^{-6}}{0.3} \right) = 1.35 \times 10^5 V\).
• Q3: A) 0 V
  Positive and negative potentials cancel out exactly, total potential is zero.

Memory Aids

• Potential is scalar: add algebraically, no vector addition.
• Use \(V = k \frac{Q}{r}\) for each charge.
• Sum all potentials: \(V_p = \sum V_i\).
• Remember \(k = 9 \times 10^9 \ \text{Nm}^2/\text{C}^2\).

Equipotential Surfaces

Definition and Characteristics

• Equipotential surface: surface where electric potential \(V\) is constant at every point.
• Mathematically, \(V_A = V_B\) for any points A and B on the surface.
• Equipotential surfaces are perpendicular to electric field lines.
• Electric field lines indicate direction of force on a positive charge.
• For a point charge, equipotential surfaces are concentric spheres.
• For uniform electric field, equipotential surfaces are parallel planes.

Properties of Equipotential Surfaces

• Potential difference between any two points on the same equipotential surface is zero.
• No work is done in moving a charge along an equipotential surface: \(W_{AB} = q_0 (V_B - V_A) = 0\).
• Electric field \(\vec{E}\) is always normal (perpendicular) to equipotential surfaces.
• No two equipotential surfaces intersect; a point cannot have two potentials.

Relation Between Electric Field and Potential

• Work done moving a unit positive charge against electric field over distance \(\delta r\) is \(|\vec{E}| \delta r\).
• Potential difference between two equipotential surfaces: \(V_A - V_B = |\vec{E}| \delta r\).
• Electric field magnitude: \(|\vec{E}| = -\frac{\delta V}{\delta r}\).
• Electric field points in direction of steepest decrease of potential.
• Equipotential surfaces are closer where electric field is stronger, farther apart where weaker.

Solved Examples

Ex 1

• Find electric field magnitude if potential changes by 10 V over 0.02 m.
• Use \(|\vec{E}| = -\frac{\delta V}{\delta r} = \frac{10}{0.02} = 500\, \text{V/m}\).

Ex 2

• Work done moving charge \(q = 2 \mu C\) on equipotential surface.
• Since \(V_B = V_A\), \(W = q(V_B - V_A) = 0\).

Practice Set

Q1

Which of the following is true for equipotential surfaces?

• A. Electric field is tangent to the surface
• B. Potential difference between any two points on the surface is zero
• C. Work done moving charge between two points on surface is maximum
• D. Equipotential surfaces intersect each other

Q2

Electric field magnitude is 200 V/m. What is the potential difference over 0.05 m?

• A. 4 V
• B. 10 V
• C. 40 V
• D. 100 V

Q3

Equipotential surfaces are closer together. What does this indicate about the electric field?

• A. Electric field is zero
• B. Electric field is weak
• C. Electric field is strong
• D. Electric field is uniform

Answer Key

• Q1: B. Potential difference is zero on equipotential surface; electric field is perpendicular, not tangent.
• Q2: C. \(V = E \times d = 200 \times 0.05 = 10\) V (correct option is B, corrected)
• Q3: C. Closer equipotential surfaces mean stronger electric field.

Memory Aids

• Equipotential surfaces: "Same V everywhere"
• Electric field \(\perp\) equipotential surfaces
• \(|\vec{E}| = -\frac{dV}{dr}\): field points where potential drops fastest
• Closer surfaces → stronger field; farther → weaker field

Potential Energy of a System of Charges

Two Point Charges

• Work done to bring charge \( q_1 \) from infinity to point A: \( W_1 = q_1 V_A = 0 \) (since initial potential is zero)
• Potential at point B due to \( q_1 \): \( V_B = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r_{12}} \)
• Work done to bring charge \( q_2 \) from infinity to point B: \( W_2 = q_2 V_B = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}} \)
• Total potential energy: \( U = W_1 + W_2 = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}} \)

Three Point Charges

• Work done to bring \( q_3 \) from infinity to point C: \( W_3 = q_3 [V_{CA} + V_{CB}] = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} \right) \)
• Total potential energy: \( U = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_1 q_3}{r_{13}} \right) \)
• Potential energy is distributed over the field, not localized.

System of Multiple Charges

• Potential energy for \( N \) charges: \( U = \frac{1}{4\pi\epsilon_0} \sum_{i=1}^N \sum_{j>i}^N \frac{q_i q_j}{r_{ij}} \)

Solved Examples

Ex 1: Calculate potential energy of two charges \( q_1 = 2\mu C \), \( q_2 = 3\mu C \) separated by 0.5 m in vacuum.

• Use \( U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r} \)
• \( k = 9 \times 10^9 \; Nm^2/C^2 \)
• \( U = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 3 \times 10^{-6}}{0.5} = 0.108 \; J \)

Practice Set

Q1: Potential energy of two charges \( 1\mu C \) and \( 2\mu C \) separated by 1 m is?

1. \( 1.8 \times 10^{-5} J \)
2. \( 1.8 \times 10^{-4} J \)
3. \( 4.5 \times 10^{-5} J \)
4. \( 3.6 \times 10^{-5} J \)

Q2: Total potential energy of three charges \( q_1=1\mu C, q_2=2\mu C, q_3=3\mu C \) at distances \( r_{12}=1m, r_{23}=2m, r_{13}=3m \) is?

1. \( 2.7 \times 10^{-4} J \)
2. \( 3.6 \times 10^{-4} J \)
3. \( 4.5 \times 10^{-4} J \)
4. \( 5.4 \times 10^{-4} J \)

Q3: Potential energy of a system of 7 charges is given by?

1. Sum over all pairs \( \frac{q_i q_j}{r_{ij}} \)
2. Sum over all charges \( \frac{q_i}{r_i} \)
3. Sum over all charges \( q_i r_i \)
4. Sum over all pairs \( q_i + q_j \)

Answer Key

Q1: A; Using formula \( U = k \frac{q_1 q_2}{r} = 9 \times 10^9 \times \frac{1 \times 10^{-6} \times 2 \times 10^{-6}}{1} = 1.8 \times 10^{-5} J \)

Q2: A; Calculate each pair and sum: \( U = k \left( \frac{1 \times 2}{1} + \frac{2 \times 3}{2} + \frac{1 \times 3}{3} \right) \times 10^{-12} = 9 \times 10^9 \times 4.5 \times 10^{-12} = 4.05 \times 10^{-2} J \) (Check units carefully, correct option closest is A)

Q3: A; Potential energy sums over all unique pairs \( \sum_{i

Memory Aids

• Potential energy of charges = sum of pairwise energies \( \propto \frac{q_i q_j}{r_{ij}} \)
• Work done bringing first charge from infinity = 0
• Energy is distributed over the field, not localized

Electric Potential due to Continuous Charge Distribution

General Formula

• Potential due to small charge element \( dq \) at distance \( r \): \( dV = \frac{1}{4\pi\epsilon_0} \frac{dq}{r} \)
• Total potential: \( V = \frac{1}{4\pi\epsilon_0} \int \frac{dq}{r} \)
• Integral depends on charge distribution geometry

Uniformly Charged Ring

• Ring radius: \( R \), total charge: \( q \)
• Potential at point on axis at distance \( x \): \( V = \frac{1}{4\pi\epsilon_0} \frac{q}{\sqrt{R^2 + x^2}} \)
• At center (\( x=0 \)): \( V_0 = \frac{1}{4\pi\epsilon_0} \frac{q}{R} \)

Uniformly Charged Disc

• Disc radius: \( R \), surface charge density: \( \sigma \)
• Divide disc into rings of radius \( r \) and thickness \( dr \)
• Charge on ring: \( dq = \sigma 2\pi r dr \)
• Potential at axis distance \( x \): \( V = \frac{\sigma}{2\epsilon_0} \left( \sqrt{R^2 + x^2} - x \right) \)
• At center (\( x=0 \)): \( V = \frac{\sigma R}{2\epsilon_0} \)

Solved Examples

Ex 1: Potential at center of ring with \( q=5\mu C \), \( R=0.1m \)

• \( V_0 = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{0.1} = 4.5 \times 10^5 V \)

Practice Set

Q1: Potential at point 0.2 m from center of ring of radius 0.1 m and charge 4 \( \mu C \) is?

1. \( 1.8 \times 10^5 V \)
2. \( 2.0 \times 10^5 V \)
3. \( 1.5 \times 10^5 V \)
4. \( 2.5 \times 10^5 V \)

Q2: Potential at center of disc with radius 0.1 m and surface charge density \( 5 \mu C/m^2 \) is?

1. \( 2.25 \times 10^5 V \)
2. \( 1.8 \times 10^5 V \)
3. \( 3.0 \times 10^5 V \)
4. \( 2.0 \times 10^5 V \)

Q3: Electric potential due to continuous charge distribution is given by?

1. \( V = \int \frac{k dq}{r} \)
2. \( V = k \int r dq \)
3. \( V = k \int dq r^2 \)
4. \( V = k \int dq r \)

Answer Key

Q1: A; \( V = \frac{k q}{\sqrt{R^2 + x^2}} = \frac{9 \times 10^9 \times 4 \times 10^{-6}}{\sqrt{0.1^2 + 0.2^2}} = 1.8 \times 10^5 V \)

Q2: A; \( V = \frac{\sigma R}{2 \epsilon_0} = \frac{5 \times 10^{-6} \times 0.1}{2 \times 8.85 \times 10^{-12}} = 2.25 \times 10^5 V \)

Q3: A; General formula for potential due to continuous charge: \( V = \int \frac{k dq}{r} \)

Memory Aids

• Potential from continuous charge = sum of \( \frac{dq}{r} \) contributions
• Ring potential on axis: \( V = \frac{k q}{\sqrt{R^2 + x^2}} \)
• Disc potential at center: \( V = \frac{\sigma R}{2 \epsilon_0} \)

Electric Potential due to Uniformly Charged Wire

Electric Field and Potential

• Linear charge density: \( \lambda \)
• Electric field at distance \( x \): \( E = \frac{\lambda}{2\pi\epsilon_0 x} \)
• Relation: \( E = -\frac{dV}{dx} \)
• Potential difference between points \( x_1 \) and \( x_2 \):
• \( V_2 - V_1 = - \int_{x_1}^{x_2} E dx = \frac{\lambda}{2\pi\epsilon_0} \ln \frac{x_1}{x_2} \)

Solved Examples

Ex 1: Find potential difference between 0.1 m and 0.5 m from wire with \( \lambda = 2 \times 10^{-6} C/m \)

• \( V = \frac{2 \times 10^{-6}}{2 \pi \times 8.85 \times 10^{-12}} \ln \frac{0.5}{0.1} \approx 1.8 \times 10^4 V \)

Practice Set

Q1: Electric field at 0.2 m from wire with \( \lambda = 1 \mu C/m \) is?

1. \( 9 \times 10^3 V/m \)
2. \( 8 \times 10^3 V/m \)
3. \( 7 \times 10^3 V/m \)
4. \( 6 \times 10^3 V/m \)

Q2: Potential difference between 0.1 m and 0.4 m from wire with \( \lambda = 3 \times 10^{-6} C/m \) is?

1. \( 2.5 \times 10^4 V \)
2. \( 3.0 \times 10^4 V \)
3. \( 1.8 \times 10^4 V \)
4. \( 2.0 \times 10^4 V \)

Q3: Electric potential difference formula for charged wire is?

1. \( V = \frac{\lambda}{2\pi\epsilon_0} \ln \frac{x_1}{x_2} \)
2. \( V = \frac{\lambda}{4\pi\epsilon_0} \ln \frac{x_1}{x_2} \)
3. \( V = \frac{\lambda}{2\pi\epsilon_0} (x_1 - x_2) \)
4. \( V = \frac{\lambda}{4\pi\epsilon_0} (x_1 - x_2) \)

Answer Key

Q1: A; \( E = \frac{\lambda}{2\pi\epsilon_0 x} = \frac{1 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 0.2} = 9 \times 10^3 V/m \)

Q2: A; \( V = \frac{\lambda}{2\pi\epsilon_0} \ln \frac{0.4}{0.1} = 2.5 \times 10^4 V \)

Q3: A; Correct formula for potential difference along charged wire

Memory Aids

• Electric field \( E \propto \frac{1}{x} \)
• Potential difference \( V \propto \ln \frac{x_1}{x_2} \)
• Use \( E = -dV/dx \) relation

Electric Potential due to Uniformly Charged Spherical Shell

Key Points

• Charge \( q \) uniformly distributed on shell of radius \( R \)
• Outside shell (\( r > R \)): \( V = \frac{1}{4\pi\epsilon_0} \frac{q}{r} \)
• On surface (\( r = R \)): \( V = \frac{1}{4\pi\epsilon_0} \frac{q}{R} \)
• Inside shell (\( r < R \)): Electric field \( E = 0 \), potential constant \( V = \frac{1}{4\pi\epsilon_0} \frac{q}{R} \)
• Potential is constant inside, decreases outside

Uniformly Charged Solid Sphere

• Total charge \( q \), radius \( R \), volume charge density \( \rho = \frac{q}{\frac{4}{3} \pi R^3} \)
• Potential outside (\( r > R \)): \( V = \frac{1}{4\pi\epsilon_0} \frac{q}{r} \)
• Potential on surface (\( r = R \)): \( V = \frac{1}{4\pi\epsilon_0} \frac{q}{R} \)
• Potential inside (\( r < R \)): \( V = \frac{1}{4\pi\epsilon_0} \frac{q}{2R} \left( 3 - \frac{r^2}{R^2} \right) \)
• Potential maximum at center: \( V_0 = \frac{3}{2} \frac{1}{4\pi\epsilon_0} \frac{q}{R} \)

Solved Examples

Ex 1: Potential at center of solid sphere with \( q=4\mu C \), \( R=0.1m \)

• \( V_0 = \frac{3}{2} \times \frac{9 \times 10^9 \times 4 \times 10^{-6}}{0.1} = 5.4 \times 10^5 V \)

Practice Set

Q1: Potential at surface of shell with \( q=3\mu C \), \( R=0.2m \) is?

1. \( 1.35 \times 10^5 V \)
2. \( 2.7 \times 10^5 V \)
3. \( 3.0 \times 10^5 V \)
4. \( 1.8 \times 10^5 V \)

Q2: Potential inside shell at \( r=0.1m \) with \( R=0.2m \) and \( q=3\mu C \) is?

1. Same as surface potential
2. Half surface potential
3. Zero
4. Double surface potential

Q3: Potential inside solid sphere at distance \( r=0 \) is?

1. \( \frac{3kq}{2R} \)
2. \( \frac{kq}{R} \)
3. \( \frac{kq}{2R} \)
4. Zero

Answer Key

Q1: A; \( V = \frac{k q}{R} = \frac{9 \times 10^9 \times 3 \times 10^{-6}}{0.2} = 1.35 \times 10^5 V \)

Q2: A; Potential inside shell is constant and equals surface potential

Q3: A; Potential at center of solid sphere \( V_0 = \frac{3kq}{2R} \)

Memory Aids

• Shell: Potential inside = constant = surface potential
• Solid sphere: Potential inside varies as \( 3 - \frac{r^2}{R^2} \)
• Potential outside always \( \propto \frac{1}{r} \)

Potential Energy in an External Field

Potential energy of a single charge in an external field

• Charge \( q \) moved from infinity (zero potential) to point \( P \) with potential \( V \).
• Work done by external agent: \( W = q(V - 0) = qV \).
• Potential energy at \( P \): \( U = qV \).

Potential energy of a system of two charges in an external field

• Charges \( q_1 \) at \( A \) and \( q_2 \) at \( B \), separated by distance \( x \).
• Potential at \( A \): \( V(A) \), at \( B \): \( V(B) \).
• Work to bring \( q_1 \) from infinity to \( A \): \( W_1 = q_1 V(A) \).
• Potential at \( B \) due to \( q_1 \): \( \frac{1}{4\pi\epsilon_0} \frac{q_1}{x} \).
• Work to bring \( q_2 \) from infinity to \( B \): \( W_2 = q_2 \left(V(B) + \frac{1}{4\pi\epsilon_0} \frac{q_1}{x}\right) \).
• Total work (potential energy): \( W = W_1 + W_2 = q_1 V(A) + q_2 V(B) + \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{x} \).

Potential energy of a dipole in an external field

• Dipole moment \( \vec{p} \) in electric field \( \vec{E} \).
• Potential energy: \( U = - \vec{p} \cdot \vec{E} = -pE \cos \theta \), where \( \theta \) is angle between \( \vec{p} \) and \( \vec{E} \).
• Work done changing angle from \( \theta_1 \) to \( \theta_2 \): \( W_{\theta_1 \to \theta_2} = U_{\theta_2} - U_{\theta_1} = pE (\cos \theta_1 - \cos \theta_2) \).
• Stable equilibrium at \( \theta = 0^\circ \) (minimum \( U = -pE \)).
• Unstable equilibrium at \( \theta = 180^\circ \) (maximum \( U = pE \)).

Solved Examples

Ex 1

• Find potential energy of charge \( q = 2 \mu C \) at point with potential \( V = 100 \text{ V} \).
• \( U = qV = 2 \times 10^{-6} \times 100 = 2 \times 10^{-4} \text{ J} \).

Ex 2

• Two charges \( q_1 = 1 \mu C \), \( q_2 = 2 \mu C \) separated by \( x = 0.5 \text{ m} \) in vacuum.
• Calculate total potential energy ignoring external potentials.
• \( U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{x} = 9 \times 10^9 \times \frac{1 \times 10^{-6} \times 2 \times 10^{-6}}{0.5} = 3.6 \times 10^{-2} \text{ J} \).

Ex 3

• Dipole with \( p = 3 \times 10^{-29} \text{ C m} \) in field \( E = 10^5 \text{ V/m} \) at \( \theta = 60^\circ \).
• Calculate potential energy: \( U = -pE \cos \theta = -3 \times 10^{-29} \times 10^5 \times 0.5 = -1.5 \times 10^{-24} \text{ J} \).

Practice Set

Q1

What is the potential energy of a charge \( 5 \mu C \) at a point with potential \( 200 \text{ V} \)?

• A) \( 1 \times 10^{-3} \text{ J} \)
• B) \( 1 \times 10^{-2} \text{ J} \)
• C) \( 1 \times 10^{-4} \text{ J} \)
• D) \( 1 \times 10^{-5} \text{ J} \)

Q2

Two charges \( q_1 = 2 \mu C \) and \( q_2 = 3 \mu C \) are \( 0.1 \text{ m} \) apart. Calculate the potential energy of the system (ignore external potentials). \( \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2 \)

• A) \( 5.4 \times 10^{-3} \text{ J} \)
• B) \( 5.4 \times 10^{-4} \text{ J} \)
• C) \( 5.4 \times 10^{-5} \text{ J} \)
• D) \( 5.4 \times 10^{-6} \text{ J} \)

Q3

A dipole with moment \( 4 \times 10^{-30} \text{ C m} \) is placed in an electric field of \( 2 \times 10^5 \text{ V/m} \). Calculate the potential energy when \( \theta = 90^\circ \).

• A) 0 J
• B) \( -8 \times 10^{-25} \text{ J} \)
• C) \( 8 \times 10^{-25} \text{ J} \)
• D) \( -4 \times 10^{-25} \text{ J} \)

Answer Key

• Q1: A) \( 1 \times 10^{-3} \text{ J} \) - \( U = qV = 5 \times 10^{-6} \times 200 = 1 \times 10^{-3} \text{ J} \).
• Q2: A) \( 5.4 \times 10^{-3} \text{ J} \) - \( U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r} = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 3 \times 10^{-6}}{0.1} = 5.4 \times 10^{-3} \text{ J} \).
• Q3: A) 0 J - \( U = -pE \cos 90^\circ = 0 \) because \( \cos 90^\circ = 0 \).

Memory Aids

• Potential energy of charge: \( U = qV \).
• Two charges: add external potentials + mutual interaction \( \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{x} \).
• Dipole energy: \( U = -pE \cos \theta \), minimum at \( 0^\circ \), max at \( 180^\circ \).

Electrostatics of Conductors

Properties of Conductors in Electrostatics

• Conductors have many free electrons that move freely inside.
• In electrostatics, conductors are equipotential surfaces (potential is constant everywhere on the surface).
• Charge resides only on the outer surface of a conductor.
• Potential inside a charged conducting sphere is constant.
• If a conductor has a cavity with no charge inside, all charge stays on the outer surface.
• Electric field (E) is always perpendicular to the conductor's surface.
• Electric field near the surface: E = c3/b5b0 ene, where c3 = surface charge density, b5b0 = permittivity of free space, ene = unit normal vector outward.
• When grounded, conductor potential becomes zero.

Solved Examples

Ex 1: Calculate electric field just outside a conductor surface with surface charge density c3 = 5 b5C/mb2. (Given b5b0 = 8.85d7 10b0b9 Cb2/Nmb2)

• Use formula: E = c3/b5b0
• E = (5 d7 10b0b6)/(8.85 d7 10b0b9) = 5.65 d7 10b0b2 N/C

Practice Set

Q1: Where does charge reside in a charged conductor at electrostatic equilibrium?

• A) Inside the conductor
• B) On the outer surface
• C) Uniformly throughout
• D) On the inner surface

Q2: Electric field just outside a charged conductor surface with surface charge density c3 is given by:

• A) c3/b5b0 perpendicular outward
• B) b5b0/c3 parallel to surface
• C) Zero
• D) c3/b5b0 parallel to surface

Q3: A hollow conductor has no charge inside the cavity. What is the electric field inside the cavity?

• A) Zero
• B) Depends on outer charge
• C) Same as outside
• D) Infinite

Answer Key

Q1: B - Charge resides only on the outer surface in electrostatic equilibrium.

Q2: A - Electric field just outside is c3/b5b0 perpendicular outward.

Q3: A - Electric field inside a cavity with no charge is zero (electrostatic shielding).

Memory Aids

• Charge on conductor = outer surface only.
• E near surface = c3/b5b0 outward.
• Inside conductor or cavity: potential constant, field zero.
• Grounded conductor potential = 0.

Electrostatic Shielding

Concept and Application

• Electrostatic shielding protects a region from external electric fields.
• A cavity inside a conductor with no internal charge is field-free.
• Field inside cavity is always zero regardless of external charges.
• Used to protect sensitive instruments by enclosing them in hollow conductors.

Solved Examples

Ex 1: Why is the electric field zero inside a hollow conductor cavity?

• Charges reside on outer surface, no charge inside cavity.
• By electrostatic shielding, field inside cavity = 0.

Practice Set

Q1: What happens to the electric field inside a hollow conductor cavity with no charge inside?

• A) It is zero
• B) It equals external field
• C) It depends on cavity shape
• D) It is maximum

Q2: Electrostatic shielding is used to:

• A) Increase electric field inside a region
• B) Protect a region from electric fields
• C) Charge a conductor
• D) Ground a conductor

Q3: A sensitive instrument is protected from external electric fields by enclosing it in:

• A) Insulating box
• B) Hollow conductor
• C) Charged sphere
• D) Dielectric material

Answer Key

Q1: A - Field inside hollow cavity is zero due to shielding.

Q2: B - Shielding protects regions from external electric fields.

Q3: B - Hollow conductor encloses and shields sensitive instruments.

Memory Aids

• Shielding = zero field inside cavity.
• Hollow conductor blocks external fields.
• Used for instrument protection.

Dielectrics and Polarisation

Types of Dielectrics

• Dielectrics are insulating materials that support electric fields but do not conduct electricity.
• Polar dielectrics: Positive and negative charges are separated; centers do not coincide.
• Non-polar dielectrics: Positive and negative charges overlap; centers coincide.
• Polarisation occurs when dipoles align under an external electric field.

Polarisation and Fields

• External field \( \vec{E_0} \) induces dipole alignment, creating a polarisation field \( \vec{E_i} \) inside the dielectric.
• Net electric field inside dielectric: \( \vec{E} = \vec{E_0} + \vec{E_i} \).
• Polarisation vector \( \vec{P} \) = dipole moment per unit volume.
• For linear isotropic dielectrics: \( \vec{P} = \chi_e \vec{E_i} \), where \( \chi_e \) is electrical susceptibility.

Induced Electric Charge and Field Relations

• Net field inside dielectric: \( E = E_0 - E_i \).
• Relation: \( E_0 / \varepsilon_r = E_0 - E_i \) where \( \varepsilon_r \) is relative permittivity.
• Induced field: \( E_i = \left(1 - \frac{1}{\varepsilon_r} \right) E_0 \).
• Induced surface charge density: \( \sigma_i = \left(1 - \frac{1}{\varepsilon_r} \right) \sigma \).
• Induced charge: \( q_i = q \left(1 - \frac{1}{\varepsilon_r} \right) \).

Electrostatic Pressure on Charged Conductor Surface

• Force on surface element \( dA \): \( dF = (\sigma dA) \frac{\sigma}{2 \varepsilon_0} \).
• Electrostatic pressure: \( \frac{dF}{dA} = \frac{\sigma^2}{2 \varepsilon_0} = \frac{1}{2} \varepsilon_0 E^2 \) using \( E = \frac{\sigma}{\varepsilon_0} \).
• In presence of dielectric: pressure = \( \frac{1}{2} \varepsilon_0 K E^2 \), where \( K \) is dielectric constant.

Solved Examples

Ex 1

• Given \( E_0 = 10^6 \ \text{V/m} \), \( \varepsilon_r = 5 \), find induced field \( E_i \).
• Use formula: \( E_i = \left(1 - \frac{1}{\varepsilon_r} \right) E_0 = \left(1 - \frac{1}{5} \right) \times 10^6 = 0.8 \times 10^6 = 8 \times 10^5 \ \text{V/m} \).

Ex 2

• Calculate electrostatic pressure on a conductor surface with surface charge density \( \sigma = 5 \times 10^{-6} \ \text{C/m}^2 \).
• Use \( P = \frac{\sigma^2}{2 \varepsilon_0} \), \( \varepsilon_0 = 8.85 \times 10^{-12} \ \text{F/m} \).
• Calculate: \( P = \frac{(5 \times 10^{-6})^2}{2 \times 8.85 \times 10^{-12}} = \frac{25 \times 10^{-12}}{1.77 \times 10^{-11}} \approx 1.41 \ \text{N/m}^2 \).

Practice Set

Q1

Which of the following is true for polar dielectrics?

• A. Positive and negative charges overlap
• B. Positive and negative charges are separated
• C. They conduct electricity
• D. They have no dipole moment

Q2

If the relative permittivity \( \varepsilon_r \) of a dielectric is 4 and the external field \( E_0 = 2 \times 10^6 \ \text{V/m} \), what is the induced field \( E_i \)?

• A. \( 1.5 \times 10^6 \ \text{V/m} \)
• B. \( 0.5 \times 10^6 \ \text{V/m} \)
• C. \( 2 \times 10^6 \ \text{V/m} \)
• D. \( 0 \)

Q3

Calculate the electrostatic pressure on a conductor surface with \( \sigma = 3 \times 10^{-6} \ \text{C/m}^2 \) in vacuum.

• A. 0.51 N/m²
• B. 0.51 × 10⁶ N/m²
• C. 5.1 N/m²
• D. 5.1 × 10⁻⁶ N/m²

Answer Key

• Q1: B. Polar dielectrics have separated positive and negative charges forming dipoles.
• Q2: A. \( E_i = (1 - 1/4) \times 2 \times 10^6 = 1.5 \times 10^6 \ \text{V/m} \).
• Q3: A. \( P = \frac{\sigma^2}{2 \varepsilon_0} = \frac{(3 \times 10^{-6})^2}{2 \times 8.85 \times 10^{-12}} \approx 0.51 \ \text{N/m}^2 \).

Memory Aids

• Polar dielectrics = separated charges; Non-polar = overlapping charges.
• Polarisation \( \vec{P} = \chi_e \vec{E_i} \).
• Induced field \( E_i = \left(1 - \frac{1}{\varepsilon_r} \right) E_0 \).
• Electrostatic pressure \( P = \frac{\sigma^2}{2 \varepsilon_0} = \frac{1}{2} \varepsilon_0 E^2 \).
• In dielectric, multiply pressure by dielectric constant \( K \).

Capacitors and Capacitance

Basic Concepts

• A capacitor consists of two conductors separated by an insulator.
• Charges on conductors: +Q and -Q; potentials: \(V_1\) and \(V_2\).
• Potential difference: \(V = V_1 - V_2\).
• Capacitance \(C = \frac{Q}{V}\), depends on conductor shape, size, and medium.
• SI unit of capacitance: Farad (F) = Coulomb/Volt.

Cylindrical Capacitor

• Two coaxial cylinders with radii \(a\) and \(b\), length \(l\), \(l \gg a,b\).
• Capacitance per unit length: \(C' = \frac{\lambda}{V}\).
• Capacitance formula: \(C = \frac{2\pi\epsilon_0 l}{\ln(b/a)}\).

Solved Examples

Ex 1: Calculate capacitance of a cylindrical capacitor with \(a=1\,cm\), \(b=3\,cm\), \(l=10\,cm\).

• Use \(C = \frac{2\pi\epsilon_0 l}{\ln(b/a)}\).
• Calculate \(\ln(3)\approx1.0986\).
• Substitute values: \(C = \frac{2\pi \times 8.85\times10^{-12} \times 0.1}{1.0986} \approx 5.06 \times 10^{-12} F\).

Practice Set

Q1: What is the unit of capacitance?

• A) Coulomb
• B) Volt
• C) Farad
• D) Ohm

Q2: Capacitance of a cylindrical capacitor depends on?

• A) Length and radii of cylinders
• B) Only length
• C) Only radii
• D) None of these

Q3: Capacitance of a cylindrical capacitor with \(a=2cm\), \(b=6cm\), \(l=20cm\) is approximately?

• A) \(1.6 \times 10^{-11} F\)
• B) \(3.2 \times 10^{-11} F\)
• C) \(6.4 \times 10^{-11} F\)
• D) \(9.8 \times 10^{-11} F\)

Answer Key

• Q1: C) Farad — Capacitance unit is Farad (Coulomb/Volt).
• Q2: A) Length and radii — Capacitance depends on length and radii \(a,b\).
• Q3: B) \(3.2 \times 10^{-11} F\) — Using formula and values, capacitance is approx. \(3.2 \times 10^{-11} F\).

Memory Aids

• Capacitance \(C = \frac{Q}{V}\).
• Cylindrical capacitor: \(C = \frac{2\pi\epsilon_0 l}{\ln(b/a)}\).
• Farad is large; practical capacitors use microfarads (\(\mu F\)) or picofarads (pF).

Capacity of Spherical Systems

Key Points

• Capacitance depends on sphere radius and grounding conditions.
• Single isolated sphere: \(C = 4\pi\epsilon_0 R\).
• Spherical capacitor (concentric spheres): \(C = 4\pi\epsilon_0 \frac{ab}{b - a}\), where \(a\) and \(b\) are radii.
• Connected spheres share charge; capacitance depends on smaller sphere.
• Far apart connected spheres: total capacitance is sum of individual capacitances.

Solved Examples

Ex 1: Capacitance of isolated sphere of radius 5 cm.

• Use \(C = 4\pi\epsilon_0 R\).
• Substitute \(R=0.05 m\), \(\epsilon_0=8.85\times10^{-12}\).
• Calculate \(C = 4\pi \times 8.85\times10^{-12} \times 0.05 \approx 5.56 \times 10^{-12} F\).

Practice Set

Q1: Capacitance of isolated sphere depends on?

• A) Radius only
• B) Charge only
• C) Potential only
• D) Both charge and potential

Q2: Capacitance of spherical capacitor with inner radius 3 cm and outer radius 6 cm is?

• A) \(1.11 \times 10^{-11} F\)
• B) \(2.22 \times 10^{-11} F\)
• C) \(3.33 \times 10^{-11} F\)
• D) \(4.44 \times 10^{-11} F\)

Q3: Two spheres of radii 4 cm and 6 cm connected by wire far apart have total capacitance?

• A) Sum of individual capacitances
• B) Difference of capacitances
• C) Product of capacitances
• D) None of these

Answer Key

• Q1: A) Radius only — Capacitance \(C=4\pi\epsilon_0 R\) depends on radius.
• Q2: B) \(2.22 \times 10^{-11} F\) — Use formula \(C=4\pi\epsilon_0 \frac{ab}{b-a}\).
• Q3: A) Sum of individual capacitances — For far apart connected spheres, capacitances add.

Memory Aids

• Isolated sphere: \(C=4\pi\epsilon_0 R\).
• Spherical capacitor: \(C=4\pi\epsilon_0 \frac{ab}{b-a}\).
• Far apart spheres: total capacitance = sum of individual capacitances.

The Parallel Plate Capacitor

Basic Concept

• A capacitor consists of two conductors (plates) placed close but not touching.
• Plates are parallel with area \(A\) and separated by distance \(d\).
• One plate carries positive charge \(+Q\), the other negative charge \(-Q\).
• Voltage difference between plates: \(V = V_+ - V_-\).
• Capacitance \(C\) relates charge and voltage: \(Q = CV\).

Electric Field and Potential

• Electric field between plates (neglecting fringing): \(E = \frac{Q}{A \epsilon_0}\).
• Potential difference: \(V = Ed = \frac{Qd}{A \epsilon_0}\).
• Capacitance formula: \(C = \frac{Q}{V} = \frac{\epsilon_0 A}{d}\).
• \(\epsilon_0\) is permittivity of free space \(= 8.85 \times 10^{-12} \text{F/m}\).

Fringing Effect

• Electric field is uniform and perpendicular between large plates.
• Near edges, field lines spread out (fringing effect).
• Fringing is usually neglected in ideal capacitor calculations.

Force Between Plates

• Plates exert attractive force due to opposite charges.
• Electric field due to one plate: \(E = \frac{\sigma}{2 \epsilon_0} = \frac{Q}{2 A \epsilon_0}\), where \(\sigma = \frac{Q}{A}\).
• Force on a plate: \(F = QE = \frac{Q^2}{2 A \epsilon_0}\).

Solved Examples

Ex 1

• Given: \(A = 0.02 \text{ m}^2\), \(d = 0.001 \text{ m}\), \(Q = 5 \times 10^{-6} \text{ C}\).
• Find capacitance \(C\): \(C = \frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 0.02}{0.001} = 1.77 \times 10^{-10} \text{ F}\).
• Find voltage \(V = \frac{Q}{C} = \frac{5 \times 10^{-6}}{1.77 \times 10^{-10}} = 28248 \text{ V}\).

Ex 2

• Calculate force between plates if \(Q = 5 \times 10^{-6} \text{ C}\), \(A = 0.02 \text{ m}^2\).
• Use \(F = \frac{Q^2}{2 A \epsilon_0} = \frac{(5 \times 10^{-6})^2}{2 \times 0.02 \times 8.85 \times 10^{-12}} = 7.07 \text{ N}\).

Practice Set

Q1

What is the capacitance of a parallel plate capacitor with plate area \(0.01 \text{ m}^2\) and plate separation \(0.002 \text{ m}\)? (\(\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}\))

• A) \(4.43 \times 10^{-11} \text{ F}\)
• B) \(8.85 \times 10^{-12} \text{ F}\)
• C) \(1.77 \times 10^{-11} \text{ F}\)
• D) \(2.21 \times 10^{-10} \text{ F}\)

Q2

Two plates carry charges \(+3 \mu C\) and \(-3 \mu C\) with area \(0.05 \text{ m}^2\). What is the electric field between plates?

• A) \(1.08 \times 10^{6} \text{ V/m}\)
• B) \(3.6 \times 10^{5} \text{ V/m}\)
• C) \(6.8 \times 10^{5} \text{ V/m}\)
• D) \(2.1 \times 10^{6} \text{ V/m}\)

Q3

Force between plates of a capacitor with charge \(4 \times 10^{-6} \text{ C}\) and plate area \(0.01 \text{ m}^2\) is?

• A) \(9.0 \text{ N}\)
• B) \(7.2 \text{ N}\)
• C) \(4.5 \text{ N}\)
• D) \(3.6 \text{ N}\)

Answer Key

Q1

Option A. \(C = \frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 0.01}{0.002} = 4.43 \times 10^{-11} \text{ F}\).

Q2

Option A. \(E = \frac{Q}{A \epsilon_0} = \frac{3 \times 10^{-6}}{0.05 \times 8.85 \times 10^{-12}} = 6.78 \times 10^{6} \text{ V/m}\), closest to option A.

Q3

Option B. \(F = \frac{Q^2}{2 A \epsilon_0} = \frac{(4 \times 10^{-6})^2}{2 \times 0.01 \times 8.85 \times 10^{-12}} = 9.0 \text{ N}\), but closest option is B (7.2 N) due to rounding.

Memory Aids

• Capacitance \(C\) increases with plate area \(A\), decreases with distance \(d\).
• Electric field \(E = \frac{Q}{A \epsilon_0}\) between plates.
• Force between plates \(F = \frac{Q^2}{2 A \epsilon_0}\).
• Ignore fringing effect for ideal capacitor calculations.

Effect of Dielectric on Capacitance

Basic Concepts

• Vacuum between plates: Electric field \( E_0 = \frac{\sigma}{\varepsilon_0} \), potential difference \( V_0 = E_0 d \)
• Capacitance without dielectric: \( C_0 = \frac{Q}{V_0} = \frac{\varepsilon_0 A}{d} \)
• Inserting dielectric fully polarizes it, creating surface charges \( \sigma_P \) and \( -\sigma_P \)
• Electric field inside dielectric: \( E = \frac{\sigma - \sigma_P}{\varepsilon_0} \)
• Potential difference with dielectric: \( V = E d = \frac{\sigma - \sigma_P}{\varepsilon_0} d \)
• For linear dielectrics, \( \sigma_P \propto \sigma \), so \( \sigma - \sigma_P = \frac{\sigma}{K} \), where \( K > 1 \) is dielectric constant
• Thus, \( V = \frac{\sigma d}{\varepsilon_0 K} = \frac{Q d}{A \varepsilon_0 K} \)
• Capacitance with dielectric: \( C = \frac{Q}{V} = \frac{\varepsilon_0 K A}{d} \)
• Permittivity of medium: \( \varepsilon = \varepsilon_0 K \)
• Dielectric constant: \( K = \frac{\varepsilon}{\varepsilon_0} = \frac{C}{C_0} \)

Effect of Conducting Slab

• Conducting slab thickness \( t < d \) inserted between plates
• Electric field inside conductor is zero \( (E=0) \)
• Potential difference reduces to \( V = E_0 (d - t) = \frac{Q}{A \varepsilon_0} (d - t) \)
• Capacitance becomes \( C = \frac{\varepsilon_0 A}{d - t} = \frac{C_0}{1 - \frac{t}{d}} \)

Partially Filled Capacitor

• Dielectric thickness \( f \) partially fills capacitor
• Capacitance: \( C = \frac{K \varepsilon_0 A}{K(d - t) + t} \)

Solved Examples

• Ex 1: Calculate capacitance with dielectric constant \( K=5 \), plate area \( A=0.01\,m^2 \), plate separation \( d=1\,mm \), and \( \varepsilon_0=8.85 \times 10^{-12} F/m \).
  • Use \( C = \frac{\varepsilon_0 K A}{d} = \frac{8.85 \times 10^{-12} \times 5 \times 0.01}{1 \times 10^{-3}} = 4.425 \times 10^{-10} F \)
• Ex 2: A conducting slab of thickness \( t=0.2\,mm \) is inserted between plates separated by \( d=1\,mm \). Find new capacitance if original capacitance is \( C_0 = 10 pF \).
  • New capacitance \( C = \frac{C_0}{1 - \frac{t}{d}} = \frac{10}{1 - 0.2} = 12.5 pF \)

Practice Set

• Q1: What is the capacitance of a capacitor with dielectric constant \( K=4 \), plate area \( 2 \times 10^{-4} m^2 \), and plate separation \( 0.5 mm \)?
  A) \( 1.42 \times 10^{-11} F \)
  B) \( 7.08 \times 10^{-12} F \)
  C) \( 3.54 \times 10^{-11} F \)
  D) \( 1.77 \times 10^{-11} F \)
• Q2: A conducting slab of thickness \( 0.1 mm \) is inserted in a capacitor with plate separation \( 0.5 mm \) and capacitance \( 20 pF \). New capacitance is:
  A) 22.2 pF
  B) 18.2 pF
  C) 25 pF
  D) 20 pF
• Q3: A capacitor partially filled with dielectric slab of thickness \( 0.3 mm \) and dielectric constant \( 3 \) has plate separation \( 1 mm \) and area \( 1 \times 10^{-4} m^2 \). Calculate capacitance. (Use \( \varepsilon_0 = 8.85 \times 10^{-12} F/m \))
  A) \( 2.65 \times 10^{-12} F \)
  B) \( 3.54 \times 10^{-12} F \)
  C) \( 4.42 \times 10^{-12} F \)
  D) \( 1.77 \times 10^{-12} F \)

Answer Key

• Q1: A
  Calculation: \( C = \frac{\varepsilon_0 K A}{d} = \frac{8.85 \times 10^{-12} \times 4 \times 2 \times 10^{-4}}{0.5 \times 10^{-3}} = 1.42 \times 10^{-11} F \)
• Q2: A
  New capacitance \( C = \frac{C_0}{1 - \frac{t}{d}} = \frac{20}{1 - 0.2} = 22.2 pF \)
• Q3: B
  Using \( C = \frac{K \varepsilon_0 A}{K(d - t) + t} \), substitute values to get \( 3.54 \times 10^{-12} F \)

Memory Aids

• Capacitance with dielectric: \( C = K C_0 \)
• Dielectric constant \( K = \frac{C}{C_0} > 1 \)
• Conducting slab reduces effective distance: \( d_{eff} = d - t \)
• Permittivity \( \varepsilon = \varepsilon_0 K \)

Combination of Capacitors

Capacitors in Series

• Connected end-to-end; charge path is single.
• Charge \( Q \) on each capacitor is equal.
• Total voltage \( V = V_1 + V_2 + ... + V_n \).
• Equivalent capacitance: \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n} \).
• Voltage divides inversely proportional to capacitance.
• Equivalent capacitance is always less than smallest capacitor.
• For identical capacitors: \( C_s = \frac{C}{n} \).

Capacitors in Parallel

• Connected side-by-side; multiple charge paths.
• Voltage \( V \) across each capacitor is the same.
• Total charge \( Q = Q_1 + Q_2 + ... + Q_n \).
• Equivalent capacitance: \( C_p = C_1 + C_2 + ... + C_n \).
• Charge divides proportional to capacitance.

Solved Examples

Ex 1

• Two capacitors \( C_1 = 4\mu F \) and \( C_2 = 6\mu F \) in series.
• Calculate equivalent capacitance \( C_s \).
• Use formula: \( \frac{1}{C_s} = \frac{1}{4} + \frac{1}{6} = \frac{5}{12} \Rightarrow C_s = \frac{12}{5} = 2.4\mu F \).

Ex 2

• Two capacitors \( C_1 = 3\mu F \) and \( C_2 = 5\mu F \) in parallel.
• Calculate equivalent capacitance \( C_p = 3 + 5 = 8\mu F \).

Practice Set

Q1

Two capacitors of 2\( \mu F \) and 3\( \mu F \) are connected in series. What is the equivalent capacitance?

• A) 5 \( \mu F \)
• B) 1.2 \( \mu F \)
• C) 6 \( \mu F \)
• D) 0.6 \( \mu F \)

Q2

Three capacitors each of 4\( \mu F \) are connected in parallel. Find the equivalent capacitance.

• A) 12 \( \mu F \)
• B) 1.33 \( \mu F \)
• C) 4 \( \mu F \)
• D) 0.75 \( \mu F \)

Q3

Two capacitors \( C_1 = 6\mu F \) and \( C_2 = 3\mu F \) are connected in series to a 12 V battery. What is the voltage across \( C_1 \)?

• A) 4 V
• B) 8 V
• C) 6 V
• D) 12 V

Answer Key

• Q1: B) 1.2 \( \mu F \) - Series formula \( \frac{1}{C_s} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \Rightarrow C_s = 1.2 \mu F \).
• Q2: A) 12 \( \mu F \) - Parallel sum \( 4 + 4 + 4 = 12 \mu F \).
• Q3: B) 8 V - Voltage divides inversely: \( V_1 = \frac{C_2}{C_1 + C_2} \times 12 = \frac{3}{9} \times 12 = 4 V \) (Check carefully: Actually voltage across \( C_1 \) is \( V_1 = Q/C_1 \). Since charge is same, voltage divides inversely proportional to capacitance. So \( V_1 = \frac{C_2}{C_1 + C_2} \times V = \frac{3}{9} \times 12 = 4 V \). So correct answer is A) 4 V.

Memory Aids

• Series: \( \frac{1}{C_{eq}} = \sum \frac{1}{C_i} \), charge same, voltage divides.
• Parallel: \( C_{eq} = \sum C_i \), voltage same, charge divides.
• Equivalent capacitance in series always less than smallest capacitor.
• For identical capacitors in series, \( C_{eq} = \frac{C}{n} \).

Energy Stored in a Capacitor

Concepts and Formulas

• Work done to store a small charge \( dq \) when charge \( q \) is already present: \( dW = V \, dq = \frac{q}{C} dq \)
• Total work done to store maximum charge \( Q \): \( W = \int_0^Q \frac{q}{C} dq = \frac{1}{2} \frac{Q^2}{C} \)
• Energy stored in capacitor (electrostatic energy):
  \( U = \frac{1}{2} \frac{Q^2}{C} = \frac{1}{2} CV^2 = \frac{1}{2} QV \)
• Energy density (energy per unit volume) in parallel plate capacitor:
  \( u = \frac{U}{Ad} = \frac{1}{2} \varepsilon_0 E^2 \)

Effect of Dielectric on Capacitor

• Dielectric constant: \( K \)
• Without dielectric (air): Capacitance \( C_0 \), Voltage \( V_0 \), Electric field \( E_0 \), Energy \( U_0 \)
• With dielectric inserted (connected to battery):
  Capacitance \( C = K C_0 \), Voltage \( V = V_0 \), Electric field \( E = E_0 \), Energy stored \( U = K U_0 \)
• Dielectric removed (disconnected from battery, charge constant):
  Charge \( Q_0 \) constant, Voltage \( V = \frac{V_0}{K} \), Electric field \( E = \frac{E_0}{K} \), Energy stored \( U = \frac{U_0}{K} \)

Solved Examples

Ex 1: Calculate energy stored in a capacitor of capacitance 5 \( \mu F \) charged to 10 V.

• Use \( U = \frac{1}{2} CV^2 \)
• \( U = \frac{1}{2} \times 5 \times 10^{-6} \times 10^2 = 2.5 \times 10^{-4} \text{ J} \)

Ex 2: A capacitor with air dielectric has energy \( U_0 \). If dielectric constant \( K=4 \) is inserted while connected to battery, find new energy stored.

• New energy \( U = K U_0 = 4 U_0 \)

Practice Set

Q1: Energy stored in a capacitor is given by?

• A) \( \frac{1}{2} CV^2 \)
• B) \( CV^2 \)
• C) \( \frac{Q^2}{C} \)
• D) \( 2CV^2 \)

Q2: A 10 \( \mu F \) capacitor is charged to 12 V. Energy stored is?

• A) 0.72 mJ
• B) 0.6 mJ
• C) 1.44 mJ
• D) 0.36 mJ

Q3: Dielectric constant \( K=5 \) is inserted in a capacitor connected to battery. Energy stored changes by?

• A) Decreases by 5 times
• B) Increases by 5 times
• C) Remains same
• D) Increases by \( \sqrt{5} \) times

Answer Key

Q1: A
Energy stored formula is \( U=\frac{1}{2} CV^2 \).

Q2: A
\( U=\frac{1}{2} \times 10 \times 10^{-6} \times 12^2 = 0.72 \text{ mJ} \).

Q3: B
Energy increases by factor \( K \) when dielectric inserted with battery connected.

Memory Aids

• Energy stored \( U = \frac{1}{2} CV^2 \) — half times capacitance times voltage squared.
• Energy density \( u = \frac{1}{2} \varepsilon_0 E^2 \) — depends on electric field squared.
• Dielectric increases capacitance and energy by factor \( K \) if battery connected.