Answer:
d.2U0
Explanation:
The capacitance of a capacitor is given by the following formula
[tex]C = \epsilon_0\frac{A}{d}[/tex]
And the electric potential energy stored in a capacitor is given by
[tex]U = \frac{Q^2}{2C}[/tex]
Applying these formula to the given variables, the initial potential energy becomes
[tex]U_0 = \frac{Q^2}{2}\frac{d}{\epsilon_0 A}[/tex]
If d becomes 2d, then potential energy becomes
[tex]U_0 = \frac{Q^2}{2}\frac{2d}{\epsilon_0 A} = 2U_0[/tex]
d. 2U₀
The electrical energy, U, stored in a capacitor is given by;
U = [tex]\frac{1}{2}[/tex][tex]\frac{Q^2}{C}[/tex] ------------------(i)
Where;
Q = Charge on the plates of the capacitor
C = capacitance of the capacitor = Aε₀ / d
A = Area of the plates of the capacitor
d = distance of separation between the plates
ε₀ = permittivity of free space.
Substitute C = Aε₀ / d into equation (i) as follows;
U = [tex]\frac{1}{2}[/tex] Q²d / Aε₀ -----------(ii)
From the question;
Electrical energy U₀, is stored when the charges on the plates are +Q and -Q.
Substitute these into equation (ii) as follows;
U₀ = [tex]\frac{1}{2}[/tex] Q²d / Aε₀ ----------------(iii)
Now, when the distance is doubled (d = 2d), the electrical energy stored becomes U₁ which is given by;
U₁ = [tex]\frac{1}{2}[/tex] Q²(2d) / Aε₀
U₁ = 2 x [tex]\frac{1}{2}[/tex] Q²d / Aε₀ [[tex]\frac{1}{2}[/tex] Q²d / Aε₀ = U₀ -------------- from equation (iii)]
U₁ = 2 x U₀
U₁ = 2U₀
Therefore, the electrical energy stored when the separation between the plates is doubled is twice the initial electrical energy stored.
