Question

A capacitor consists of a set of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery that maintains a constant potential difference V across the plates. If the separation between the plates is doubled, the electrical energy stored in the capacitor will be
a.doubled
b. unchanged
c. quadrupled
d. quartered
e. halved

Answer 1

Answer:

(e) halved

Explanation:

The electrical enery (E) stored in a capacitor is related to its capacitance (C) and potential difference (V) as follows;

E = $\frac{1}{2}$ x C x $V^{2}$   ------------------------(i)

Also, the capacitance (C) of a capacitor consisting of parallel plates is related to the area (A) of the plates and distance (d) between the plates as follows;

C = A x ε₀ / d    ------------------------(ii)

Where;

ε₀ is the permittivity of free space.

Substituting equation (ii) into equation (i) gives;

E = $\frac{1}{2}$ x A x ε₀ / d x $V^{2}$  --------------------(iii)

From equation(iii)

When the potential difference (V) is constant, then the electrical energy (E) stored is inversely proportional to the distance between the plates. i.e

E = k / d   ----------------(iv)

Where;

k = proportionality constant = $\frac{1}{2}$ x A x ε₀ x $V^{2}$ (which is the product of all constants)

Therefore from equation (iv);

=> E₁ x d₁ = E₂ x d₂   ---------------------------(v)

Where;

E₁ and E₂ are the initial and final values of the electrical energy stored.

d₁ and d₂ are the initial and final values of the distance between the plates.

So, when the distance is doubled, i.e.

d₂ = 2 x d₁

Substitute the value of d₂ into equation (v) to give;

=> E₁ x d₁ = 2 x d₁ x E₂

Divide through by d₁ to give;

=> E₁ = 2 x E₂

Make E₂ subject of the formula

=> E₂ = $\frac{1}{2}$ x E₁

Therefore, the electrical energy stored in the capacitor will be halved.