(e) halved
The electrical enery (E) stored in a capacitor is related to its capacitance (C) and potential difference (V) as follows;
E = [tex]\frac{1}{2}[/tex] x C x [tex]V^{2}[/tex] ------------------------(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 = [tex]\frac{1}{2}[/tex] x A x ε₀ / d x [tex]V^{2}[/tex] --------------------(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 = [tex]\frac{1}{2}[/tex] x A x ε₀ x [tex]V^{2}[/tex] (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 = [tex]\frac{1}{2}[/tex] x E₁
Therefore, the electrical energy stored in the capacitor will be halved.
