Electrostatic Potential and Capacitance - Class 12 Physics

Class 12 Physics | Unit I — Electrostatics

Chapter 2: Electrostatic Potential and Capacitance

Electric Potential • Potential Due to Dipole • Equipotential Surfaces • Capacitors • Energy • Dielectrics

1. Electric Potential

Electric Potential (V): Work done per unit positive test charge in bringing it from infinity to the given point, against the electric force. V = W_∞→P / q₀. Unit: Volt (V = J/C). Scalar quantity.

Due to a point charge Q at distance r:
V = kQ/r = Q/(4πε₀r)
k = 9×10⁹ N·m²/C², ε₀ = 8.85×10⁻¹² C²/N·m²

Due to a system of charges: V = k∑Q_i/r_i (algebraic sum — scalar addition)

Potential is a scalar: can be positive, negative, or zero.
V = 0 at infinity (reference). V > 0 for +Q, V < 0 for −Q.
Relation to electric field: E = −dV/dr (E is negative gradient of V). E points from high V to low V.
Potential difference: V_A − V_B = W_B→A/q (work done in moving positive charge from B to A).
For a conductor in electrostatic equilibrium: entire surface at same potential (equipotential). E = 0 inside conductor. E ⊥ surface outside.

1.1 Relationship Between E and V

In one dimension: E = −dV/dx
In general: E_x = −∂V/∂x, E_y = −∂V/∂y, E_z = −∂V/∂z
For uniform field: V_A − V_B = Ed (d = distance along field direction)

⚠️ NEET: V = kQ/r (point charge). V is scalar. E = −dV/dr. At the surface of sphere: V = kQ/R, E = kQ/R². Inside sphere: V = kQ/R (constant), E = 0. Work done = q(V_A−V_B). Path-independent (conservative force).

2. Potential Due to an Electric Dipole

Electric Dipole: Two equal and opposite charges (+q and −q) separated by distance 2a. Dipole moment p = q(2a), directed from −q to +q. Unit: C·m.

Potential at point P (distance r, angle θ with dipole axis):
V = kp cosθ / r² (for r >> a)

Special cases:
• On axial line (θ = 0°): V_axial = kp/r² (maximum, positive)
• On equatorial line (θ = 90°): V_eq = 0
• V ∝ 1/r² for dipole (vs 1/r for point charge)

2.1 Electric Field Due to Dipole

On axial line (end-on): E_axial = 2kp/r³ (along p direction)
On equatorial line (broad-side): E_eq = kp/r³ (opposite to p direction)
E_axial = 2 × E_eq at same distance r

2.2 Torque on Dipole in Uniform Electric Field

τ = pE sinθ = p × E
Potential energy: U = −pE cosθ = −p·E
• θ = 0°: stable equilibrium (U = −pE, minimum)
• θ = 90°: τ = pE (maximum)
• θ = 180°: unstable equilibrium (U = +pE, maximum)

⚠️ NEET (2013, 2016, 2019, 2021): V_axial = kp/r²; V_eq = 0. E_axial = 2kp/r³ (along p); E_eq = kp/r³ (opposite to p). Torque τ = pE sinθ. At equatorial: V=0 but E≠0. Dipole moment direction: −q to +q.

3. Equipotential Surfaces

Equipotential Surface: A surface over which electric potential is the same (constant) at every point. No work is done in moving a charge on an equipotential surface.

E field ⊥ equipotential surface always (if E had component along surface, work would be done → contradiction).
Equipotential surfaces never intersect each other.
Closer equipotential lines → stronger electric field (E = −dV/dr; smaller dr, larger E).
For point charge: equipotential surfaces are concentric spheres.
For uniform E field: equipotential surfaces are planes ⊥ to E.
For electric dipole: complex curved surfaces (not spheres or planes).
Work done in moving charge on equipotential surface = 0 (ΔV = 0 → W = qΔV = 0).

4. Potential Energy of a System of Charges

Two charges: U = kq₁q₂/r₁₂
Three charges: U = k[q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₁q₃/r₁₃]
Work done in bringing q₂ from ∞ to r: W = kq₁q₂/r
Energy stored = work done in assembling the configuration.

U > 0: like charges (energy stored = repulsion work done against force).
U < 0: unlike charges (energy released as charges approach).
Electric PE is a property of the system, not individual charges.

5. Capacitors and Capacitance

Capacitance (C): The ability of a conductor to store charge. C = Q/V. Unit: Farad (F). 1 F = 1 C/V. A capacitor consists of two conductors (plates) separated by an insulating medium (dielectric or vacuum).

5.1 Parallel Plate Capacitor (vacuum)

C₀ = ε₀A / d
A = area of each plate (m²), d = separation between plates (m)
Electric field between plates: E = σ/ε₀ = Q/(ε₀A) = V/d (uniform field)
σ = Q/A = surface charge density (C/m²)

5.2 Effect of Dielectric

With dielectric (dielectric constant K = ε_r):
C = KC₀ = Kε₀A / d = ε₀ε_rA / d
Electric field with dielectric: E = E₀/K (reduced)
Dielectric constant K ≥ 1 (always). Vacuum: K = 1.

Polarisation: In dielectric, molecules align with E field → induced dipoles create opposing (bound) charges on surface → reduces E₀ to E₀/K → capacitance increases.
Polar dielectrics (e.g., H₂O): have permanent dipole moments; align in E field.
Non-polar dielectrics (e.g., O₂): develop induced dipoles in E field.
If battery remains connected and slab inserted: V constant, Q increases, C increases.
If battery disconnected and slab inserted: Q constant, V decreases, C increases, E decreases.

5.3 Capacitors in Series and Parallel

Series: 1/C_eff = 1/C₁ + 1/C₂ + 1/C₃ (same charge, voltage divides)
For 2: C_eff = C₁C₂/(C₁+C₂)

Parallel: C_eff = C₁ + C₂ + C₃ (same voltage, charge divides)
(Opposite rules to resistors!)

5.4 Energy Stored in a Capacitor

U = QV/2 = Q²/(2C) = CV²/2
Energy density (energy per unit volume between plates):
u = ½ε₀E² (J/m³)

When two charged capacitors are connected: charge redistributes until both reach same V. Energy is lost (as heat in connecting wire) unless ideal conditions.
Common potential: V_common = (C₁V₁ + C₂V₂) / (C₁ + C₂)

⚠️ NEET (2014, 2016, 2018, 2020, 2022): C = ε₀A/d. With dielectric K: C = Kε₀A/d. Series: 1/C = Σ1/C_i (like parallel R). Parallel: C = ΣC_i (like series R). Energy = CV²/2. Energy density = ½ε₀E². Common V = (C₁V₁+C₂V₂)/(C₁+C₂). Energy lost when connecting two capacitors.

6. Van de Graaff Generator

Van de Graaff Generator: A device that can produce very high electrostatic potentials (of the order of millions of volts) by continuously accumulating charge on a large metallic sphere. Used in nuclear physics research to accelerate charged particles.

Principle: (a) Charge is always distributed on outer surface of conductor. (b) Electrostatic action of sharp points (corona discharge / brush discharge). (c) The potential of a conducting sphere is determined by its radius (large sphere → higher V for same Q).
Construction: Large hollow metallic sphere mounted on insulating column. A moving belt of insulating material runs from base to sphere. Two sharp-tipped metallic brushes — one sprays charge onto belt, other collects it onto sphere.
Working: Motor drives belt; lower brush sprays +ve charge onto belt via corona discharge (connected to +ve terminal of supply). Belt carries charge to interior of sphere. Upper brush collects charge from belt and transfers to sphere. As Q builds up on sphere, V rises (V = Q/(4πε₀R)).
Max potential: V_max reached when electric field at surface = breakdown field of surrounding gas. Air breakdown: ~3×10⁶ V/m. Inert gas (SF₆) used to increase V_max.
Use: Accelerate protons, alpha particles to bombard nuclei; artificial nuclear reactions.

7. Capacitor vs Resistor Combinations

Combination	Resistors	Capacitors
Series	R_eff = R₁+R₂+... (adds up)	1/C_eff = 1/C₁+1/C₂+... (reciprocal adds)
Parallel	1/R_eff = 1/R₁+1/R₂+... (reciprocal adds)	C_eff = C₁+C₂+... (adds up)
Same in series	Same current I	Same charge Q
Same in parallel	Same voltage V	Same voltage V

Key memory trick: Capacitor combinations are exactly opposite to resistor combinations — series capacitors behave like parallel resistors, and vice versa.

🎓 NEET Previous Year Questions

Q1. [NEET 2022] On the equatorial line of an electric dipole, the electric potential is:

Answer: Zero V_eq = kp cos90°/r² = 0. On equatorial line, θ = 90°, so V = 0 for all distances. However, E ≠ 0 on equatorial line (E_eq = kp/r³, directed opposite to p).

Q2. [NEET 2021] A parallel plate capacitor with air has capacitance C. If a dielectric slab (K=3) fills completely the gap, new capacitance is:

Answer C' = KC = 3C. C = Kε₀A/d = K × (original C). Inserting dielectric always increases capacitance by factor K.

Q3. [NEET 2020] Three capacitors 2 µF, 3 µF, 6 µF are connected in series. Effective capacitance:

Answer 1/C = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1. C = 1 µF.

Q4. [NEET 2019] The energy stored in a 50 µF capacitor charged to 100 V is:

Answer U = CV²/2 = (50×10⁻⁶×100²)/2 = (50×10⁻⁶×10000)/2 = 0.5/2 = 0.25 J.

Q5. [NEET 2018] The electric field inside a conductor in electrostatic equilibrium is:

Answer: Zero In electrostatic equilibrium, free charges redistribute until no net force acts on any charge → E_inside = 0. Consequence: all charge resides on outer surface; entire conductor is an equipotential.

Q6. [NEET 2017] Two capacitors (C₁=4µF, C₂=6µF) are connected in series across 200V. Charge on each and voltage across each:

Answer C_eff = 4×6/(4+6) = 2.4 µF. Q = C_effV = 2.4×200 = 480 µC (same on each in series).
V₁ = Q/C₁ = 480/4 = 120 V; V₂ = Q/C₂ = 480/6 = 80 V. Check: 120+80=200V ✓

💡 Rapid Revision — Key Formulas

V = kQ/r | E = −dV/dr | V at surface of sphere = kQ/R; inside = kQ/R (constant)
Dipole: V_axial = kp/r²; V_eq = 0; E_axial = 2kp/r³; E_eq = kp/r³
Torque on dipole: τ = pE sinθ; U = −pE cosθ
C = ε₀A/d (vacuum) | C = Kε₀A/d (with dielectric)
Series: 1/C_eff = ∑1/C_i | Parallel: C_eff = ∑C_i
Energy: U = CV²/2 = Q²/2C = QV/2 | Energy density = ½ε₀E²
Common V = (C₁V₁+C₂V₂)/(C₁+C₂) | Energy lost on connection

Numericals - Electrostatic Potential and Capacitance

CLASS 12 PHYSICS | NCERT SOLUTIONS

Chapter 2 — Electrostatic Potential and Capacitance

25 NCERT Exercise & Exemplar Questions — Step-by-Step Solutions

📝 NCERT Exercise Questions (2.1 – 2.16)

3 MarksQ1 (Ex 2.1). Two charges 5×10⁻⁸ C and −3×10⁻⁸ C are located 16 cm apart. Find (a) point on line joining charges where V = 0. (b) Another such point.

✓ Solution
Q₁ = 5×10⁻⁸ C, Q₂ = −3×10⁻⁸ C, d = 0.16 m

(a) Between the charges (let x from Q₁):
V = kQ₁/x + kQ₂/(0.16−x) = 0
5/x = 3/(0.16−x) ⇒ 5(0.16−x) = 3x ⇒ 0.80 = 8x
x = 0.10 m = 10 cm from Q₁ (between charges)

(b) Outside (beyond Q₂), let point be at distance y from Q₂:
5/(0.16+y) = 3/y ⇒ 5y = 3(0.16+y) ⇒ 2y = 0.48 ⇒ y = 0.24 m
x = 40 cm from Q₁ (24 cm beyond Q₂)

3 MarksQ2 (Ex 2.2). A regular hexagon of side 10 cm has charge 5×10⁻⁶ C at each of its six vertices. Find potential at centre.

✓ Solution
All six charges are at equal distance R = 10 cm = 0.10 m from centre.
Total potential (scalar sum):

V = 6 × kQ/R = 6 × 9×10⁹ × 5×10⁻⁶ / 0.10

V = 6 × 9×10⁹ × 5×10⁻⁵ = 6 × 4.5×10⁵ = 2.7×10⁶ V = 2.7 MV

3 MarksQ3 (Ex 2.3). Two charges 2×10⁻⁶ C and −2×10⁻⁶ C are at (−1 cm, 0) and (+1 cm, 0). Calculate electric field at (a) midpoint between them, (b) point 6 cm from midpoint on axial line.

✓ Solution
p = q×2a = 2×10⁻⁶×0.02 = 4×10⁻⁸ C·m

(a) Midpoint (r = 0.01 m from each charge, equatorial point for E field):
Each charge contributes E = kQ/r² toward −q (both fields in same direction).
E = 2×kQ/r² = 2×9×10⁹×2×10⁻⁶/(0.01)² = 2×9×2×10³/10⁻⁴ = 3.6×10⁸ N/C (along +x to −q direction)

(b) Axial point (x = 6 cm = 0.06 m from midpoint):

E_axial = 2kp/r³ (for r >> a) = 2×9×10⁹×4×10⁻⁸/(0.06)³

= (720×10¹) / (2.16×10⁻⁴) = 720/(2.16×10⁻⁴) ≈ 3.33×10⁶ N/C (along dipole axis direction)

3 MarksQ4 (Ex 2.4). An electric dipole with dipole moment 4×10⁻⁹ C·m is aligned at 30° with a uniform E field of 5×10⁴ N/C. Find torque on dipole.

✓ Solution
p = 4×10⁻⁹ C·m, E = 5×10⁴ N/C, θ = 30°

τ = pE sinθ = 4×10⁻⁹ × 5×10⁴ × sin30°

τ = 4×10⁻⁹ × 5×10⁴ × 0.5 = 20×10⁻⁵ × 0.5
τ = 10⁻⁴ N·m

3 MarksQ5 (Ex 2.5). A parallel plate capacitor has plates of area 5×10⁻³ m², separated by 0.88 mm air gap. Find (a) capacitance, (b) charge for 100 V, (c) E between plates.

✓ Solution
A = 5×10⁻³ m², d = 0.88 mm = 8.8×10⁻⁴ m

(a) Capacitance:

C = ε₀A/d = (8.85×10⁻¹²×5×10⁻³) / (8.8×10⁻⁴) = 44.25×10⁻¹⁵ / 8.8×10⁻⁴

C = 5.027×10⁻¹¹ F = 50.3 pF

(b) Charge at 100 V:

Q = CV = 50.3×10⁻¹² × 100 = 5.03 nC

(c) Electric field:

E = V/d = 100 / (8.8×10⁻⁴) = 1.14×10⁵ N/C

3 MarksQ6 (Ex 2.6). A parallel plate capacitor is charged to V = 100 V, then disconnected from battery. A dielectric slab (K = 3) is inserted filling the gap. Find new: C, V, E, Q, and U. (Initial C = 200 pF)

✓ Solution
Initial: C₀ = 200 pF, V₀ = 100 V, Q₀ = 200×100 = 20,000 pC = 20 nC
Battery disconnected → Q remains constant = 20 nC

New C = KC₀ = 3×200 = 600 pF
New V = Q/C = 20000/600 = 33.3 V (decreased by K)
New E = V/d = E₀/K = 100/d × 1/3 → E = E₀/3 (decreased)
Q = 20 nC (unchanged, battery disconnected)
Initial U = Q²/(2C₀) = (20000)²/(2×200) = 10⁶ pJ
New U = Q²/(2C) = (20000)²/(2×600) = U₀/3 (energy decreases — energy goes into polarising dielectric)

3 MarksQ7 (Ex 2.7). A 12 pF capacitor is connected to a 50 V battery. Find energy stored. If another 12 pF capacitor is connected in parallel, find new energy stored.

✓ Solution
Initial: C = 12 pF, V = 50 V

U₁ = CV²/2 = (12×10⁻¹²×50²)/2 = (12×10⁻¹²×2500)/2 = 1.5×10⁻⁸ J = 15 nJ

With second 12 pF in parallel (battery still connected):
C_new = 12+12 = 24 pF, V = 50V (battery keeps voltage constant)

U₂ = C_newV²/2 = (24×10⁻¹²×2500)/2 = 3.0×10⁻⁸ J = 30 nJ

3 MarksQ8 (Ex 2.8). Three capacitors C₁=2µF, C₂=3µF, C₃=4µF are connected in series to a 100 V supply. Find: (a) C_eff, (b) charge on each, (c) voltage across each.

✓ Solution
(a) Series C_eff:

1/C = 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12

C_eff = 12/13 µF ≈ 0.923 µF

(b) Charge (same on all in series):

Q = C_effV = (12/13) × 100 ≈ 92.3 µC

(c) Voltages:
V₁ = Q/C₁ = 92.3/2 = 46.15 V
V₂ = Q/C₂ = 92.3/3 = 30.77 V
V₃ = Q/C₃ = 92.3/4 = 23.08 V
Check: 46.15+30.77+23.08 = 100 V ✓

3 MarksQ9 (Ex 2.9). A 600 pF capacitor is charged by a 200 V supply. It is then disconnected and connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost?

✓ Solution
C = 600 pF, V = 200 V
Initial energy: U_i = CV²/2 = (600×10⁻¹²×200²)/2 = 60×10⁻⁷/2 = 1.2×10⁻⁵ J
After connection: Common V_c = Q_total/(C₁+C₂) = (600×200)/(600+600) = 100 V
Final energy: U_f = (C₁+C₂)V_c²/2 = (1200×10⁻¹²×100²)/2 = 6×10⁻⁶ J

Energy lost = U_i − U_f = 1.2×10⁻⁵ − 6×10⁻⁶ = 6×10⁻⁶ J = 6 µJ

This energy is dissipated as heat in connecting wires.

3 MarksQ10 (Ex 2.10). In a parallel plate capacitor with air, the electric field is 10⁵ N/C when V = 100 V. A dielectric slab (K = 4) of same dimensions is inserted. Find new E, C ratio, and V (battery still connected).

✓ Solution
Battery connected → V remains constant = 100 V
d = V/E₀ = 100/10⁵ = 10⁻³ m = 1 mm
New E = V/d = 100/10⁻³ = 10⁵ N/C (same — V and d unchanged!)
Note: With battery connected, E = V/d = unchanged. But D = Kε₀E (displacement) increases.
C_new/C₀ = K = 4; New charge Q = KC₀V = 4 times original; Charge increases by factor 4.

3 MarksQ11. Five capacitors of 2 µF each are connected: two in series (A), and three in parallel (B), and then A and B are connected in series across 100 V. Find charge on each capacitor.

✓ Solution
Group A (2 in series): C_A = 2×2/(2+2) = 1 µF
Group B (3 in parallel): C_B = 2+2+2 = 6 µF
A and B in series: 1/C = 1/1 + 1/6 = 7/6; C_total = 6/7 µF
Total Q = C_totalV = (6/7)×100 ≈ 85.7 µC
Same Q on A and B (series): Q_A = 85.7 µC; Q_B = 85.7 µC
Each cap in A: Q = 85.7 µC (series → same Q)
V across B = Q_B/C_B = 85.7/6 = 14.3 V. Each cap in B: Q = 2×14.3 = 28.6 µC

3 MarksQ12. Calculate the potential at distance 3 cm from a charge of 10µC and at 5 cm from another charge of −8µC. What is the work done to bring a charge of 2×10⁻⁹ C from infinity to a point equidistant (4 cm) from both charges?

✓ Solution
V at equidistant point (r = 0.04 m from each):
V = k(Q₁/r + Q₂/r) = (9×10⁹/0.04)(10×10⁻⁶ − 8×10⁻⁶)
V = (9×10⁹/0.04)(2×10⁻⁶) = 225×10⁹×2×10⁻⁶ = 450×10³ = 4.5×10⁵ V
Work done: W = qV = 2×10⁻⁹×4.5×10⁵ = 9×10⁻⁴ J = 0.9 mJ

3 MarksQ13. A spherical conductor of radius 12 cm has a charge of 1.6×10⁻⁷ C. Find (a) V at surface, (b) E at surface, (c) V at 18 cm, (d) E inside at 4 cm.

✓ Solution
R = 12 cm = 0.12 m, Q = 1.6×10⁻⁷ C

(a) V at surface:

V = kQ/R = 9×10⁹×1.6×10⁻⁷/0.12 = 1.44×10³/0.12 = 12,000 V = 12 kV

(b) E at surface:

E = kQ/R² = 12000/0.12 = 10⁵ N/C (or E = V/R)

(c) V at r = 18 cm: V = kQ/r = 9×10⁹×1.6×10⁻⁷/0.18 = 1.44×10³/0.18 = 8 kV

(d) E inside at 4 cm: E = 0 (electric field inside a conductor = 0 in electrostatics)

3 MarksQ14. The plates of a parallel plate capacitor of area 90 cm² are separated by 2.5 mm. It is charged with a 400 V battery. Find energy stored initially. If battery is disconnected and plate separation doubled, find new energy stored and loss/gain.

✓ Solution
A = 90 cm² = 90×10⁻⁴ m², d₁ = 2.5×10⁻³ m, V = 400 V
C₁ = ε₀A/d₁ = 8.85×10⁻¹²×90×10⁻⁴/(2.5×10⁻³) = 79.65×10⁻¹⁶/2.5×10⁻³ = 31.86 pF ≈ 32 pF

U₁ = C₁V²/2 = 32×10⁻¹²×(400)²/2 = 32×10⁻¹²×80000 = 2.56×10⁻⁶ J = 2.56 µJ

After disconnecting and doubling d: C₂ = C₁/2 = 16 pF; Q = const = C₁V = 32×400 = 12800 pC
U₂ = Q²/(2C₂) = (12800×10⁻¹²)²/(2×16×10⁻¹²) = 5.12 µJ (energy doubled! — work done against attractive force to separate plates)

3 MarksQ15. Two identical metal spheres A (charge +3µC) and B (charge −5µC) are brought into contact and then separated. Find charge on each and work done.

✓ Solution
On contact: charges redistribute equally.
Total charge = +3 + (−5) = −2 µC
Charge on each after separation: −2/2 = −1 µC each
They share equally since identical spheres.

3 MarksQ16. A 10 µF capacitor is charged to 12 V. It is then disconnected and connected to an uncharged 5 µF capacitor. Find the common voltage, charge on each, and energy lost.

✓ Solution
C₁ = 10 µF, V₁ = 12V, C₂ = 5µF (uncharged)
Q_initial = C₁V₁ = 10×12 = 120 µC

V_c = (C₁V₁+C₂×0)/(C₁+C₂) = 120/(10+5) = 120/15 = 8 V

Q₁ = C₁V_c = 10×8 = 80 µC; Q₂ = C₂V_c = 5×8 = 40 µC
U_i = C₁V₁²/2 = 10×144/2 = 720 µJ; U_f = (C₁+C₂)V_c²/2 = 15×64/2 = 480 µJ
Energy lost = 720 − 480 = 240 µJ

🌟 Additional / Exemplar Questions (Q17 – Q25)

3 MarksQ17. If four identical capacitors C are connected: three in series and one in parallel with this series combination, across V volts — find total capacitance and energy.

✓ Solution
Three in series: C_series = C/3
One C in parallel with C/3: C_total = C + C/3 = 4C/3
Energy: U = C_totalV²/2 = 2CV²/3

3 MarksQ18. Two point charges +4 µC and −2 µC are placed 20 cm apart. Find the point on the line joining them where the electric potential is zero.

✓ Solution
Let zero potential be at distance x from +4µC (between charges):
k(4)/x + k(−2)/(0.20−x) = 0 ⇒ 4/x = 2/(0.20−x)
4(0.20−x) = 2x ⇒ 0.80 = 6x ⇒ x = 0.133 m ≈ 13.3 cm from +4µC
(Also: one point at x = 40 cm from +4µC outside, beyond −2µC)

3 MarksQ19. A conducting sphere of radius 10 cm is charged to 1000 V. Find charge and surface charge density.

✓ Solution
R = 0.10 m, V = 1000 V
V = kQ/R ⇒ Q = VR/k = 1000×0.10/(9×10⁹) = 100/9×10⁹
Q = 1.11×10⁻⁸ C = 11.1 nC
σ = Q/(4πR²) = 1.11×10⁻⁸/(4π×0.01) = 1.11×10⁻⁸/0.1257 = 8.84×10⁻⁸ C/m²

3 MarksQ20. Find the equivalent capacitance between A and B for a circuit where C₁ = 6 µF and C₂ = 3 µF connected in series, and C₃ = 4 µF in parallel with the series combination.

✓ Solution
C₁ and C₂ in series: C₁₂ = C₁C₂/(C₁+C₂) = 6×3/(6+3) = 18/9 = 2 µF
C₁₂ in parallel with C₃: C_eff = 2+4 = 6 µF

3 MarksQ21. The potential difference between the plates of a parallel-plate capacitor is 100 V when they are 2 mm apart. Find: (a) E between plates, (b) surface charge density on plates (air gap).

✓ Solution
V = 100 V, d = 2 mm = 2×10⁻³ m
(a) E = V/d = 100/(2×10⁻³) = 5×10⁴ N/C
(b) σ = ε₀E = 8.85×10⁻¹²×5×10⁴ = 4.425×10⁻⁷ C/m²

3 MarksQ22. Derive the expression for energy stored in a capacitor and energy density. Explain where the energy is stored physically.

✓ Solution
Small work dW = V dQ to add charge dQ to capacitor at potential V = Q/C:
dW = (Q/C)dQ ⇒ Total W = ∫dW = ∫₀^Q(Q/C)dQ = Q²/(2C)

U = Q²/(2C) = CV²/2 = QV/2

Energy density: Volume between plates = Ad. U = ½CV² = ½(ε₀A/d)(Ed)² = ½ε₀E²(Ad)
u = U/Vol = ½ε₀E² J/m³
Physical location: Energy is stored in the electric field existing in the space between the plates (not in the charges or plates themselves). This is a profound result applicable to all electromagnetic fields.

3 MarksQ23. Two capacitors (C₁=3µF Q=90µC, C₂=9µF uncharged) are connected with same polarity plates together. Find final charge and voltage on each.

✓ Solution
Total charge = 90+0 = 90 µC. Final common V: V_c = Q/(C₁+C₂) = 90/12 = 7.5 V
Q₁ = C₁V_c = 3×7.5 = 22.5 µC; Q₂ = 9×7.5 = 67.5 µC; Both at 7.5 V.

3 MarksQ24. A 4µF capacitor is charged to 200V and a 2µF to 300V. They are connected with plates of opposite polarity together (+ to −). Find final common voltage.

✓ Solution
Q₁ = 4×200 = 800 µC (let this be +ve)
Q₂ = 2×300 = 600 µC (now opposite polarity: −ve in this arrangement)
Net Q = 800 − 600 = 200 µC

V_c = Q_net/(C₁+C₂) = 200/(4+2) = 33.3 V

3 MarksQ25. Explain why (a) the electric potential inside a conductor is the same as on its surface, and (b) equipotential surfaces near a point charge are spherical while for a uniform field they are planar.

✓ Solution
(a) Conductor is equipotential:
Inside conductor in electrostatic equilibrium, E = 0. Work done in moving charge from one point to another inside = W = ∫E·dl = 0 (since E=0). So V_A−V_B = 0 for any two interior points. All interior points and the surface are at the same potential. The entire conducting body is one equipotential volume.

(b) Shape of equipotential surfaces:
For point charge: V = kQ/r = constant ⇒ r = constant ⇒ surface is a sphere (all points at same r from charge have same V). For uniform E (say along x-axis): V = V₀ − Ex = constant ⇒ x = constant ⇒ planes perpendicular to x-axis (the field direction). Equipotential surfaces are always ⊥ to field lines.

✍ Score Guide — 25 Questions
NCERT Ex 2.1–2.16: 3 marks each — 48 marks | Additional Q17–Q25: 3 marks each — 27 marks | Grand Total: 75 marks

Formula Capsule - Electrostatic Potential and Capacitance

CLASS 12 PHYSICS | FORMULA CAPSULE

Electrostatic Potential and Capacitance

Chapter 2 — Complete Formula Sheet & High-Yield Facts for NEET/JEE

📅 Section A — Electric Potential

Quantity	Formula	Key Note
Potential (point charge)	V = kQ/r	Scalar; + for +Q, − for −Q
System of charges	V = k∑Q_i/r_i	Algebraic (scalar) sum
Sphere surface	V = kQ/R	Same as point charge at R
Inside sphere	V = kQ/R = constant	V same everywhere inside; E = 0
Relation E and V	E = −dV/dr	E points from high V to low V
Work done	W = q(V_A−V_B)	Path-independent (conservative)
PE of two charges	U = kq₁q₂/r	+ve for like, −ve for unlike

📅 Section B — Electric Dipole

Quantity	Formula	Key Note
Dipole moment	p = q×2a	Direction: −q to +q; Unit: C·m
V on axial line	V = kp cosθ/r² ⇒ kp/r² at θ=0	Max, positive
V on equatorial line	V = 0	θ=90°; zero always
E on axial line	E_axial = 2kp/r³	Along p direction
E on equatorial line	E_eq = kp/r³	Opposite to p direction
Torque on dipole	τ = pE sinθ	Max at θ=90°; zero at θ=0°, 180°
PE of dipole	U = −pE cosθ	Min (stable) at θ=0; Max at θ=180°

📅 Section C — Capacitors

Quantity	Formula	Key Note
Capacitance	C = Q/V	Unit: Farad = C/V
Parallel plate (air)	C₀ = ε₀A/d	A=area, d=separation
With dielectric K	C = Kε₀A/d = KC₀	K ≥ 1; always increases C
E between plates	E = V/d = σ/ε₀	Uniform field
Capacitors in series	1/C_eff = 1/C₁+1/C₂+...	Same Q; voltage divides
Capacitors in parallel	C_eff = C₁+C₂+...	Same V; charge divides
Energy stored	U = CV²/2 = Q²/2C = QV/2	All 3 forms equivalent
Energy density	u = ½ε₀E²	Energy in electric field per m³
Common voltage	V_c = (C₁V₁+C₂V₂)/(C₁+C₂)	After connecting two capacitors

🧠 Memory Tricks

“C combos are OPPOSITE to R” Capacitors in SERIES → reciprocals add (like R in parallel). Capacitors in PARALLEL → values add (like R in series). Easy check: series C is always less than smallest C.

“V_eq = 0, E_eq ≠ 0” On equatorial line of dipole: potential = 0 (symmetric cancellation of +q and −q). But electric field ≠ 0 (fields add, not cancel). Classic NEET trap.

“Battery On vs Off — Dielectric” Battery ON (V fixed): C ↑, Q ↑, E same (V/d unchanged). Battery OFF (Q fixed): C ↑, V ↓ (=Q/KC), E ↓ (=E₀/K), U ↓. Energy goes into polarising dielectric.

“Inside conductor: E=0, V=const” No E field inside → no work done moving charge → V is same everywhere inside and on surface. Charge resides only on outer surface. Conductor is always an equipotential body.

“Energy lost when connecting capacitors” When charged capacitor connects to uncharged one: charge flows → current → heat in wire. Energy is always LOST (even ideal wire: spark). Final U always less than initial U.

“E_axial = 2E_eq” Axial field = 2kp/r³; Equatorial field = kp/r³. Axial is DOUBLE the equatorial at same r. Both ∝ 1/r³ (falls faster than point charge 1/r²).

🔢 Critical Values

k = 9×10⁹ N·m²/C² ε₀ = 8.85×10⁻¹² C²/N·m² V_equatorial = 0 always E inside conductor = 0 E_axial = 2E_eq Energy density = ½ε₀E² Dielectric: C increases by K Series C: 1/C = ∑1/C_i

❌ Common Mistakes to Avoid

V is scalar, E is vector: To find potential of system, simply add V_i = kQ_i/r_i algebraically. For field, add vectors (components).
Equatorial V=0 but E≠0: Both fields from +q and −q contribute to E in same direction on equatorial line but cancel for V.
Series capacitor: C < smallest C_i: Unlike series resistors (R > largest R_i), series capacitors give smaller C than any individual.
Dielectric with battery on: V fixed → E = V/d unchanged! Q increases. Mistake: thinking E reduces when dielectric inserted with battery on.
U = CV²/2 vs U = Q²/2C: Use CV²/2 when V is given/constant. Use Q²/2C when Q is conserved (battery disconnected). Don't mix them up wrongly.

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