Question

A capacitor has a potential difference of V applied across its plates. If the potential difference across its plates is increased to 2V, the energy stored in the capacitor is: A) reduced to half
B) doubles
C) reduced to one fourth
D) quadruples (four times bigger)
E) None of the above

Answer 1

Answer:

Option B. doubles

Explanation:

This can be explained by the definition of capacitance that charge on the capacitor will remain constant irrespective of the voltage applied.

This can be given as:

Q ∝ V or Q = CV

where,

Q = charge

V = Voltage

C = Capacitance

So, when V is doubled, C shoul reduce to half to main constant charge on the capacitor:

V' = 2V

C' = $\frac{1}{2}C$

Also , Energy stored in a capacitor, E is given by:

E = $\frac{1}{2}CV^{2}$                             (1)

Now, when

V' = 2V

C' = $\frac{1}{2}C$

Using eqn (1):

E' = $\frac{1}{2}C'V'^{2}$

Energy, E' = $\frac{1}{4}C(2V)^{2}$

E' = $CV^{2}$ = 2E