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

(e) halved

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

E = x C x   ------------------------(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 = x A x ε₀ / d x  --------------------(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 = x A x ε₀ x (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 = x E₁

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

answer is c

explanation: