Chapter 2: Electrostatic Potential and Capacitance

The electric potential decreasing along the direction of the electric field is given as E = dV/dr

The electric potential decreases when the path from the charged conductor is taken to the uncharged conductor along the direction of the electric field.

This continues when another uncharged conductor is considered to the infinity lowering the potential even further.

This shows that the uncharged body is at intermediate potential, and the charged body is at infinity potential.

U is the potential energy of a point charge q which is placed at potential V, U = qV

A negatively charged particle is placed at the axis of the ring with charge Q

Let a be the radius of the ring

The electric potential at the axial distance is given as

The potential energy, U is given as

When z = infinity, U = 0.

When z = 0, U is given as

Point P is considered to be at a distance z from the centre of the ring. The charge dq is at a distance z from point P.

V is written as

Therefore, the net potential is given as

The equation of the equipotential for an infinite cylinder of radius r₀ with linear charge density λ is given as:

r = r₀e 2πε₀/ λ [V(r) – V(r₀)]

The potential at point P due to charges is given as

Vp = 1/4πε₀ q/r₁ + 1/4 πε₀ (-q)/r₂

The net electric potential at this point is zero,

Therefore, r₁ = r₂

We know that

To solve the two equations, we require the equation in-plane x = 0, which is a y-z plane.

Since the capacitors are connected in parallel, the potential difference across the capacitors is the same. The final voltage is assumed to be U. C is the capacitance of the capacitor without dielectric, then the charge is given as Q₁ = CU

The initial charge is given as

Q₀ = CU₀

The conversion of charges is

Q₀ = Q₁ + Q₂

CU₀ = CU + αCU₂

αU₂ + U – U₀ = 0

Solving the equation, we get U = 6V

When the conducting disc is placed at the centre of the bottom plate, the potential of the disc is equal to the potential of the plate. The electric field on the disc is given as

E = V/d

The charge q’ is transferred to the disc, which is given as

q’ = -ε₀ V/d πr²

The force acting on the disc is

F = ε₀ V²/d² πr²

Therefore, V = square root of mdg/ π ε₀r²

There are three charges in the system. The potential energy of the system is equal to the sum of the PE of each pair.

U = 1/4 π ε₀ {q_dq_d/r – q_uq_d/r – q_uq_d/r}

Substituting the values we get,

U = 5.11 10^-4

Following are the charges on the metal sphere before contact

Q₁ = σ.4πR²

Q₂ = σ.4π(2R)² = 4Q₁

Q₁’ and Q₂’ are the charges on the metal sphere after contact

Q₁’ + Q₂’ = Q₁ + Q₂ = 5Q₁

When the metal spheres are in contact, the following is the potentials acquired by them

Q₁’ = Q₂’/2

Solving the equations we

σ₁ = 5 σ/3

σ₂ = 5 σ/6

When K₁ is closed, and K₂ is open, the capacitors C₁ and C₂ are connected in series with the battery

Therefore, the charge in capacitors C₁ and C₂ are

Q₁ = Q₂ = q = (C₁C₂/C₁+C₂)E = 18μC

When the capacitors C₂ and C₃ are placed in parallel,

C₂V’ + C₃V’ = Q₂

V’ = Q₂/C₂ + C₃ = 3V

Therefore,

Q₂’ = 3CV’ = 9 μC

Q₃’ = 3CV’ = μC

Q₁’ = 18 μC