An ideal parallel-plate capacitor consists of a set of two parallel plates of area AAA separated by a very small distance ddd. When this capacitor is connected to a battery that maintains a constant potential difference between the plates, the energy stored in the capacitor is UUU0. If the separation between the plates is doubled, how much energy is stored in the capacitor?

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An ideal parallel-plate capacitor consists of a set of two parallel plates of area AAA separated by a very small distance ddd. When this capacitor is connected to a battery that maintains a constant potential difference between the plates, the energy stored in the capacitor is UUU0. If the separation between the plates is doubled, how much energy is stored in the capacitor?

a)2U0

b)U0/2

c)4U0

d)U0/4

e)U0

Answer : ( b )

Explanation:

Firstly we calculate the capacitance of a parallel plate capacitor

C =

where e = permittivity of free space,

d = distance between plates

A = Area of plates

Note: for a parallel plate capacitor when the distance between the plates is increased by a factor of 2 ( doubled ) the capacitance of the capacitor is decreased by factor of 2

hence the new capacitance would be

C₂ =

secondly we calculate the energy stored in a parallel plate capacitor

U₀ =  ( equation 1 )

where c = capacitance of the capacitor

           V = voltage

Note: for a parallel plate capacitor when the capacitance is reduced by a factor of 2 the energy stored in the capacitor also decreases by a factor of 2 as well

hence the new energy stored would be

U₂ = * ( )   (equation 2 )

substitute  with U₀ in equation 2

The new energy becomes

U₂ = * U  =