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

Option B. doubles

Step-by-step 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' =
`(1)/(2)C`

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

E =
`(1)/(2)CV^(2)` (1)

Now, when

V' = 2V

C' =
`(1)/(2)C`

Using eqn (1):

E' =
`(1)/(2)C'V'^(2)`

Energy, E' =
`(1)/(4)C(2V)^(2)`

E' =
`CV^(2)` = 2E