Question

When a certain capacitor carries charge of magnitude Q on each of its plates, it stores energy Ep. In order to store twice as much energy, how much charge should it have on its plates

Answer 1

Answer:

2Q

Explanation:

When a capacitor carries some certain charge, the energy stored in the capacitor is its electric potential energy E. The magnitude of this potential energy is given by;

E  = $\frac{1}{2}qV$            ------------(i)

Where;

q = charge between the plates of the capacitor

V = potential difference between the plates of the capacitor

From the question;

q = Q

E = Ep

Therefore, equation (i) becomes;

Ep = $\frac{1}{2} QV$              ----------------(ii)

Make V subject of the formula in equation (ii)

V = $\frac{2E_{p}}{Q}$

Now, when the energy is doubled i.e E = 2Ep, equation (i) becomes;

2Ep = $\frac{1}{2}qV$

Substitute the value of V into the equation above;

2Ep = $\frac{1}{2}$($q *\frac{2E_{p}}{Q}$)

Solve for q;

$2E_{p}$ = $\frac{2qE_p}{2Q}$

$2E_{p}$ = $\frac{qE_p}{Q}$

$q = 2Q$

Therefore, the charge, when the energy stored is twice the originally stored energy, is twice the original charge. i.e 2Q