Electrostatic Potential and Capacitance

1. Electrostatic Potential

Definition: Work done by an external force in bringing a unit positive charge from infinity to that point (without acceleration).

V = W_ext / q
SI Unit: Volt (V) or Joule/Coulomb (J/C).
It is a Scalar quantity.

Potential due to Point Charge

V = kQ / r

(V ∝ 1/r)

Potential due to Dipole

At distance r and angle θ:

V = kp cosθ / r²

Axial (θ=0°): V = kp/r² (Max)
Equatorial (θ90°): V = 0 (Min)

(Asked in NEET 2021)

2. Equipotential Surfaces

A surface with constant value of potential at all points.

Properties

No work is done in moving a charge (W = qΔV = 0).
Electric field is always normal (90°) to the surface.
Two surfaces never intersect.
Closer surfaces → Stronger Electric Field.

3. Potential Energy

Work done in assembling a system of charges.

U = k q₁ q₂ / r

(U is Negative for attraction, Positive for repulsion)

Dipole in External Field

Work done in rotating dipole:

W = pE(cosθ₁ - cosθ₂)
U = -p ⋅ E = -pE cosθ

(Asked in NEET 2023)

4. Capacitance

Ability of a conductor to store charge.

Q = CV
SI Unit: Farad (F)

Parallel Plate Capacitor

C = Kε₀A / d

Combinations

Type	Formula	Constant Quantity
Series	1/C_eq = Σ 1/C_i	Charge (Q)
Parallel	C_eq = Σ C_i	Potential (V)

(Asked in NEET 2022)

5. Energy & Dielectrics

Energy Stored

U = ½ CV² = Q² / 2C
Energy Density: u = ½ ε₀ E²

Common Potential

When two capacitors are connected in parallel:

V = (C₁V₁ + C₂V₂) / (C₁ + C₂)

Energy is lost as heat during redistribution.

(Asked in NEET 2018)

Numericals: Potential & Capacitance

Q1. Calculate the work done in moving a charge of 4 C from a point at 220 V to a point at 230 V.

Solution:
W = q(V_final - V_initial)
Given q = 4 C, V_i = 220 V, V_f = 230 V.
W = 4 × (230 - 220)
W = 4 × 10 = 40 Joules.

Q2. A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 80 V. What is the potential at the center of the sphere?

Solution:
For a conducting sphere, electric field inside is zero.
Therefore, potential is constant throughout the volume and equals the surface potential.
V_center = V_surface = 80 V.

Q3. A parallel plate capacitor has air capacitance 8 pF. What will be the capacitance if distance is halved and space is filled with dielectric K = 6?

Solution:
Initial C₀ = ε₀A / d = 8 pF.
New d' = d/2, K = 6.
C' = Kε₀A / d' = 6 ε₀A / (d/2)
C' = 12 (ε₀A / d) = 12 × 8 pF
C' = 96 pF.

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

Solution:
U = ½ CV²
U = 0.5 × (12 × 10^-12) × (50)²
U = 6 × 10^-12 × 2500
U = 15000 × 10^-12 = 1.5 × 10^-8 J.

Q5. A 600 pF capacitor charged by 200 V is disconnected and connected in parallel to an uncharged 600 pF capacitor. Calculate common potential.

Solution:
By conservation of charge:
V = (C₁V₁ + C₂V₂) / (C₁ + C₂)
V = (600 × 200 + 0) / (600 + 600)
V = 120000 / 1200 = 100 V.

Q6. Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in series. Find the equivalent capacitance.

Solution:
1/C_eq = 1/2 + 1/3 + 1/4
LCM is 12. Numerator = 6 + 4 + 3 = 13.
1/C_eq = 13/12
C_eq = 12/13 pF ≈ 0.92 pF.

Q7. 27 identical drops of mercury are charged to a potential of 10 V each. They drop to form a big drop. Find potential of the big drop.

Solution:
Radius R = n^1/3r = 3r.
Charge Q = nq = 27q.
Potential V' = kQ/R = k(27q)/(3r) = 9 (kq/r)
V' = 9V = 9 × 10 = 90 V.

Q8. If potential V = 4x² volts, find the electric field at point (1m, 0, 2m).

Solution:
E = - ∇V. Only x-component exists.
E_x = - d/dx(4x²) = -8x.
At x = 1, E = -8(1) = -8 V/m.
E = -8 î V/m.

Q9. Dipole p = 4 × 10^-9 Cm is at 30° to field E = 5 × 10⁴ N/C. Calculate potential energy.

Solution:
U = -pE cosθ
U = -(4 × 10^-9)(5 × 10⁴) cos(30°)
U = -20 × 10^-5 × 0.866
U = -17.32 × 10^-5 = -1.73 × 10^-4 J.

Q10. A capacitor is charged and battery disconnected. A dielectric slab is interested. What happens to energy stored?

Solution:
Battery disconnected → Charge Q is constant.
Capacitance increases: C' = KC.
Energy U = Q² / 2C.
New Energy U' = Q² / 2(KC) = U / K.
Since K > 1, Energy decreases.

Important Formulae

1. Electrostatic Potential

Point Charge:

V = kQ / r

Dipole (General Point):

V = kp cosθ / r²

2. Potential Energy & Work

Two Point Charges:

U = k q₁ q₂ / r

Work Done:

W = q(V_B - V_A)

Dipole in Field:

U = -pE cosθ

3. Relation between E and V

E = - dV / dr

For uniform field: E = V / d

4. Capacitance

Parallel Plate:

C = Kε₀A / d

Energy Stored:

U = ½ CV² = Q² / 2C

Common Potential:

V = (C₁V₁ + C₂V₂) / (C₁ + C₂)

20 NEET Golden Facts

1. Scalar Nature: Electric potential is a scalar quantity, whereas electric field is a vector.
2. High to Low: Electric field lines always flow from higher potential to lower potential.
3. Ref. Potential: Potential at infinity is arbitrarily taken as zero. Earth's potential is also zero.
4. Electron Volt: Is a unit of Energy. 1 eV = 1.6 × 10^-19 Joules.
5. Equipotential Work: No work is done in moving a test charge over an equipotential surface.
6. Conductor Surface: Surface of a conductor is always an equipotential surface. Field is normal to it.
7. Inside Conductor: Electric field is zero inside a hollow charged conductor, but potential is constant (same as surface).
8. Infinite K: Dielectric constant K for metals (conductors) is infinity.
9. C Factors: Capacitance depends ONLY on geometry (shape, size, separation) and medium, NOT on Q or V.
10. Dielectric Effect: Inserting a dielectric slab completely always increases capacitance by factor K.
11. Coalescing Drops (V): When n identical drops coalesce, new potential V' = n^2/3 V.
12. Coalescing Drops (C): New capacitance of coalesced big drop C' = n^1/3 C.
13. Corona Discharge: Charge density and E-field is strongest near sharp points (Radius of curvature is small).
14. Series Cap: V ∝ 1/C. Smallest capacitor has largest potential drop. Charge is same.
15. Parallel Cap: Only charge distributes Q ∝ C. Potential difference is same.
16. Battery Disconnected: Charge Q remains constant. Inserting dielectric decreases V, decreases E, decreases U.
17. Battery Connected: Potential V remains constant. Inserting dielectric increases Q, increases C, increases U.
18. 50% Rule: Energy stored is ½QV. Work by battery is QV. So 50% energy is always dissipated as heat during charging.
19. Stable Dipole: Stable equilibrium when dipole moment is parallel to Field (θ=0). PE is minimum (-pE).
20. Energy Density: Energy per unit volume in any electric field (capacitor or free space) is ½ε₀E².

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