Question

An ideal parallel-plate capacitor consists of a set of two parallel plates of area A separated by a very small distance d. When the capacitor plates carry charges +Q and -Q, the capacitor stores energy U0. If the separation between the plates is doubled, how much electrical energy is stored in the capacitor?
a.4U0
b.U0/4
c.U0
d.2U0
e.U0/2

Answer 1

Answer:

d.2U0

Explanation:

The capacitance of a capacitor is given by the following formula

$C = \epsilon_0\frac{A}{d}$

And the electric potential energy stored in a capacitor is given by

$U = \frac{Q^2}{2C}$

Applying these formula to the given variables, the initial potential energy becomes

$U_0 = \frac{Q^2}{2}\frac{d}{\epsilon_0 A}$

If d becomes 2d, then potential energy becomes

$U_0 = \frac{Q^2}{2}\frac{2d}{\epsilon_0 A} = 2U_0$

Answer 2

Answer:

d. 2U₀

Explanation:

The electrical energy, U, stored in a capacitor is given by;

U = $\frac{1}{2}$$\frac{Q^2}{C}$            ------------------(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 = $\frac{1}{2}$ 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 = $\frac{1}{2}$ Q²d / Aε₀                ----------------(iii)

Now, when the distance is doubled (d = 2d), the electrical energy stored becomes U₁ which is given by;

U₁ = $\frac{1}{2}$ Q²(2d) / Aε₀

U₁ = 2 x $\frac{1}{2}$ Q²d / Aε₀          [$\frac{1}{2}$ 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.