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' = [tex]\frac{1}{2}C[/tex]
Also , Energy stored in a capacitor, E is given by:
E = [tex]\frac{1}{2}CV^{2}[/tex] (1)
Now, when
V' = 2V
C' = [tex]\frac{1}{2}C[/tex]
Using eqn (1):
E' = [tex]\frac{1}{2}C'V'^{2}[/tex]
Energy, E' = [tex]\frac{1}{4}C(2V)^{2}[/tex]
E' = [tex]CV^{2}[/tex] = 2E
