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Energy of a parallel plate capacitor with charge Q and potential difference V can be calculated in two ways:

1)Work done in bringing charge dq from infinty to the capacitor when there is potential V= q/C on the positive plate and V= 0 on the negative plate is dW = (q/C)dq

integrrating from 0 to Q gives W = (1/2)Q

2)Using eqn.(2.43) of Griffith's Introduction to electrodynamics , 3 ed.,

W= ## \frac 1 2\int_Γ ρV \, dΓ =\frac 1 2 \int_{q=0}^Q V \, dq = \frac 1 2 \int \frac q C \, dq = \frac 1 2 \frac {Q^2} {2C} = \frac 1 4 \frac {Q^2} {C}##,

where Γ is the volume over which the integration has to be done

How to decide which method is right?

Can anyone tell me why there is a factor of (1/2) in the above equation?

1)Work done in bringing charge dq from infinty to the capacitor when there is potential V= q/C on the positive plate and V= 0 on the negative plate is dW = (q/C)dq

integrrating from 0 to Q gives W = (1/2)Q

^{2}/C2)Using eqn.(2.43) of Griffith's Introduction to electrodynamics , 3 ed.,

W= ## \frac 1 2\int_Γ ρV \, dΓ =\frac 1 2 \int_{q=0}^Q V \, dq = \frac 1 2 \int \frac q C \, dq = \frac 1 2 \frac {Q^2} {2C} = \frac 1 4 \frac {Q^2} {C}##,

where Γ is the volume over which the integration has to be done

How to decide which method is right?

Can anyone tell me why there is a factor of (1/2) in the above equation?

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