Class 12 Physics · Chapter 2

Electrostatic Potential
and Capacitance

Master concepts of electric potential, equipotential surfaces, capacitors & energy — essential for the AAI ATC exam.

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7 Subtopics
Full Coverage
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21+ MCQs
Numerical Practice
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Key Formulas
Quick Reference
2.1
Electrostatic Potential Energy Conservative force · Work done by external agency

What is Electrostatic Potential Energy?

When an external force moves a charge q from point R to point P against the repulsive electrostatic force, work is done and stored as potential energy. Since the Coulomb force is conservative, this work depends only on the initial and final positions — not on the path taken.

The potential energy difference between two points is defined as:
ΔU = UP − UR = WRP (work done by external force)

📐 Key Formula
ΔU = UP − UR = WRP
U(r) = q·V(r)  —  potential energy of charge q at position r
U = 1/(4πε₀) · q₁q₂/r₁₂  —  for a system of two charges
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Insert Diagram: Test charge q moving from R to P against repulsive force of Q (Fig 2.1 NCERT)

AAI ATC Tip: The Coulomb force is conservative — like gravitation. This means potential energy is path-independent. Remember: work done against electric force = gain in PE.
Practice MCQs — Section 2.1
Q1 A charge of 2 × 10⁻⁶ C is moved from a point where potential is 100 V to a point where potential is 200 V. The work done by the external force is:
1 × 10⁻⁴ J
2 × 10⁻⁴ J
4 × 10⁻⁴ J
5 × 10⁻⁴ J
W = q·ΔV = 2×10⁻⁶ × (200−100) = 2×10⁻⁶ × 100 = 2 × 10⁻⁴ J
Q2 Two charges q₁ = 3 μC and q₂ = −3 μC are 0.1 m apart. The electrostatic potential energy of the system is (k = 9×10⁹ Nm²/C²):
+0.81 J
+0.27 J
−0.81 J
−0.27 J
U = kq₁q₂/r = 9×10⁹ × 3×10⁻⁶ × (−3×10⁻⁶) / 0.1 = −9×10⁻¹ × 9×10⁻¹ = −0.81 J (attractive, unlike charges)
Q3 Work done in assembling three charges q each at the corners of an equilateral triangle of side a is:
kq²/a
2kq²/a
3kq²/a
kq²/3a
Three pairs, each contributing kq²/a: U = 3 × kq²/a = 3kq²/a
2.2
Electrostatic Potential (V) Work per unit charge · Point charge potential · Dipole potential

Defining Electric Potential

Electrostatic potential V at a point is the work done by an external force in bringing a unit positive charge (without acceleration) from infinity to that point:

V = W/q (work done per unit positive charge from ∞ to the point)

Only the potential difference (VP − VR) is physically significant; the absolute value is arbitrary.

📐 Key Formulas
V(r) = 1/(4πε₀) · Q/r    — potential due to point charge Q
V = 1/(4πε₀) · p·cosθ/r²   — potential due to dipole
V = 1/(4πε₀) · Σ(qᵢ/rᵢP)   — superposition for system of charges
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Insert Diagram: Variation of V (∝ 1/r) and E (∝ 1/r²) with distance r from a point charge (Fig 2.4 NCERT)

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Insert Diagram: Potential due to an electric dipole — geometry with r₁, r₂, θ (Fig 2.5 NCERT)

Remember for AAI ATC: Dipole potential ∝ 1/r² (falls faster than point charge 1/r). Equatorial plane potential of a dipole = 0.
Practice MCQs — Section 2.2
Q4 A charge Q = 4 × 10⁻⁷ C is placed at the origin. The electric potential at a point 9 cm away is (k = 9×10⁹):
2 × 10⁴ V
4 × 10⁴ V
8 × 10⁴ V
1 × 10⁴ V
V = kQ/r = 9×10⁹ × 4×10⁻⁷ / 0.09 = 36000/0.09 × 0.01 = 4 × 10⁴ V
Q5 A dipole of moment p = 5 × 10⁻⁸ C·m. The potential at a point r = 0.5 m on the axial line (θ = 0°) is (k = 9×10⁹):
900 V
450 V
1800 V
3600 V
V = kp·cosθ/r² = 9×10⁹ × 5×10⁻⁸ × cos0° / (0.5)² = 450/0.25 = 1800 V
Q6 Two charges 3×10⁻⁸ C and −2×10⁻⁸ C are 15 cm apart. The potential is zero at how many points on the line joining them?
Only 1 (between them)
Only 1 (outside)
2 points
0 points
V = 0 at x = 9 cm (between charges) and x = 45 cm (outside, on the side of negative charge). 2 points in total.
2.3
Equipotential Surfaces E ⊥ equipotential · Relation between E and V

What is an Equipotential Surface?

A surface on which the electric potential is constant at every point. No work is done in moving a charge along an equipotential surface. The electric field E is always perpendicular to equipotential surfaces and points in the direction of steepest potential decrease.

Relation Between E and V

The magnitude of the electric field is related to the rate of change of potential:

|E| = −dV/dl (negative gradient of potential)

Electric field points from higher to lower potential; its magnitude = potential drop per unit distance normal to the equipotential surface.

📐 Key Relations
|E| = −dV/dl    — E is negative gradient of V
Equipotential of point charge: concentric spheres
Equipotential for uniform E: planes ⊥ to E
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Insert Diagram: Equipotential surfaces for (a) single point charge — concentric spheres, (b) dipole, (c) uniform field — parallel planes (Figs 2.9, 2.10, 2.11 NCERT)

Practice MCQs — Section 2.3
Q7 Two equipotential surfaces are 0.02 m apart and the potential difference between them is 8 V. The electric field between them is:
0.16 V/m
4 V/m
400 V/m
160 V/m
E = ΔV/Δl = 8/0.02 = 400 V/m
Q8 A charge is moved along an equipotential surface by a force F over distance 0.5 m. The work done is:
F × 0.5 J
0 J
2F J
0.5F J
On an equipotential surface ΔV = 0, so W = qΔV = 0 J
Q9 The electric field in a region is 500 V/m. The potential difference between two points 4 cm apart along the field direction is:
125 V
20 V
2000 V
12.5 V
ΔV = E × d = 500 × 0.04 = 20 V
2.4
Electrostatics of Conductors E = 0 inside · Surface charge · Shielding

Key Properties of Conductors in Electrostatics

Six important results govern the electrostatics of conductors:

#Property
1E = 0 inside a conductor in static equilibrium
2E is normal to the surface at every point outside
3Excess charge resides only on the outer surface
4Potential is constant throughout the volume and on the surface
5E at surface = σ/ε₀ (outward normal direction)
6Electrostatic shielding — cavity inside conductor has E = 0
📐 Surface Field Formula
E = (σ/ε₀) n̂    — field just outside a charged conductor
σ = surface charge density, n̂ = outward unit normal
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Insert Diagram: Electrostatic properties of a conductor — E=0 inside, charge on surface, field normal outside (Fig 2.19 NCERT)

Practice MCQs — Section 2.4
Q10 A spherical conductor of radius 0.12 m has surface charge density 2×10⁻⁵ C/m². The electric field just outside the surface is (ε₀ = 8.85×10⁻¹² C²/Nm²):
1130 V/m
2260 V/m
565 V/m
4520 V/m
E = σ/ε₀ = 2×10⁻⁵ / 8.85×10⁻¹² ≈ 2260 V/m
Q11 A spherical conductor of radius 12 cm has charge 1.6×10⁻⁷ C. The electric field at a point 18 cm from its centre is (k = 9×10⁹):
2.5×10⁴ V/m
4.4×10⁴ V/m
8.8×10⁴ V/m
1.2×10⁴ V/m
Outside: E = kQ/r² = 9×10⁹ × 1.6×10⁻⁷ / (0.18)² = 1440/0.0324 ≈ 4.4 × 10⁴ V/m
Q12 A charge of 10 μC is placed inside a closed conducting shell. The charge on the outer surface of the shell is:
0 μC
−10 μC
+10 μC
5 μC
Inner surface gets −10 μC (induced). By charge conservation, outer surface = +10 μC.
2.5
Dielectrics and Polarisation Polar & non-polar molecules · Susceptibility · Dielectric constant K

What are Dielectrics?

Dielectrics are non-conducting substances. In an external electric field, non-polar molecules develop induced dipole moments (charge displacement). Polar molecules tend to align their permanent dipoles with the field. This collective effect is called polarisation.

Polarisation P = ε₀χeE, where χe is the electric susceptibility of the dielectric.

📐 Key Concepts
P = ε₀χeE    — polarisation (dipole moment per unit volume)
K = ε/ε₀    — dielectric constant (K > 1)
ε = ε₀K    — permittivity of medium
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Insert Diagram: Polar and non-polar molecules, and polarisation in external field (Figs 2.21, 2.22 NCERT)

Practice MCQs — Section 2.5
Q13 A dielectric has susceptibility χe = 3. Its dielectric constant K is:
3
4
2
0.33
K = 1 + χe = 1 + 3 = 4
Q14 The polarisation P in a dielectric is 8×10⁻⁸ C/m² when E = 2×10³ V/m. The electric susceptibility χe is (ε₀ = 8.85×10⁻¹²):
≈ 4.5
≈ 0.22
≈ 9.0
≈ 2.25
χe = P/(ε₀E) = 8×10⁻⁸ / (8.85×10⁻¹² × 2×10³) = 8×10⁻⁸ / 1.77×10⁻⁸ ≈ 4.5
Q15 A mole of a substance with permanent dipole moment 10⁻²⁹ C·m is polarised in field 10⁶ V/m. If field direction changes 60°, heat released is (NA = 6×10²³):
6 J
−3 J
3 J
9 J
p = N×p₀ = 6×10⁻⁶ C·m; Uᵢ = −pEcos0° = −6 J; Uf = −pEcos60° = −3 J; ΔU = 3 J released as heat = 3 J
2.6
Capacitors and Capacitance Parallel plate · Dielectric effect · Series & Parallel combinations

What is a Capacitor?

A capacitor is a system of two conductors separated by an insulator (dielectric). Capacitance C = Q/V depends only on geometry (shape, size, separation) and the dielectric material. SI unit: Farad (F) = C/V.

📐 Key Formulas
C = Q/V    — definition of capacitance
C₀ = ε₀A/d    — parallel plate (vacuum)
C = Kε₀A/d = KC₀    — with dielectric K
Series: 1/C = 1/C₁ + 1/C₂ + …
Parallel: C = C₁ + C₂ + …
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Insert Diagram: Parallel plate capacitor with and without dielectric, showing field E₀ and E (Figs 2.25, 2.23 NCERT)

Practice MCQs — Section 2.6
Q16 A parallel plate capacitor has area A = 6×10⁻³ m², separation d = 3 mm, connected to 100 V. Capacitance C is (ε₀ = 8.85×10⁻¹²):
1.77×10⁻¹¹ F
1.77×10⁻¹¹ F (17.7 pF)
8.85×10⁻¹² F
5.31×10⁻¹¹ F
C = ε₀A/d = 8.85×10⁻¹² × 6×10⁻³ / 3×10⁻³ = 8.85×10⁻¹² × 2 = 17.7 pF
Q17 Three capacitors each of 9 pF are connected in series to a 120 V supply. The potential difference across each capacitor is:
360 V
90 V
40 V
120 V
In series, total C = 9/3 = 3 pF. Same charge on each. V per cap = 120/3 = 40 V
Q18 A parallel plate capacitor (C = 8 pF) has its plate separation halved and dielectric of K = 6 inserted. The new capacitance is:
48 pF
96 pF
24 pF
16 pF
C' = K × (ε₀A/(d/2)) = 2K × C₀ = 2 × 6 × 8 = 96 pF
2.7
Energy Stored in a Capacitor Energy formulas · Energy density of electric field

Energy Stored in a Capacitor

When a capacitor is charged, work is done and stored as electrostatic potential energy in the electric field between the plates. This energy can be expressed in three equivalent forms.

📐 Energy Formulas
U = Q²/(2C) = (1/2)CV² = (1/2)QV    — energy stored
u = (1/2)ε₀E²    — energy density (J/m³) in electric field
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Insert Diagram: Energy stored in capacitor — step-by-step charging process and energy in E field between plates (Fig 2.30 NCERT)

Practice MCQs — Section 2.7
Q19 A 900 pF capacitor is charged by a 100 V battery. Energy stored is:
9 × 10⁻⁶ J
4.5 × 10⁻⁶ J
4.5 × 10⁻⁸ J
1.8 × 10⁻⁵ J
U = (1/2)CV² = 0.5 × 900×10⁻¹² × (100)² = 0.5 × 9×10⁻⁶ = 4.5 × 10⁻⁶ J
Q20 A 12 pF capacitor is connected to a 50 V battery. Energy stored is:
3 × 10⁻⁸ J
6 × 10⁻⁹ J
1.5 × 10⁻⁸ J
6 × 10⁻⁸ J
U = (1/2)CV² = 0.5 × 12×10⁻¹² × 2500 = 0.5 × 3×10⁻⁸ = 1.5 × 10⁻⁸ J
Q21 The electric field between the plates of a parallel plate capacitor is 5×10⁴ V/m. Energy density in this field is (ε₀ = 8.85×10⁻¹²):
0.011 J/m³
0.011 J/m³ (≈11.1×10⁻³)
22 J/m³
1.1×10⁻³ J/m³
u = (1/2)ε₀E² = 0.5 × 8.85×10⁻¹² × (5×10⁴)² = 0.5 × 8.85×10⁻¹² × 25×10⁸ = ≈ 11.1×10⁻³ J/m³

📋 Chapter Summary — Quick Revision

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