Energy Stored in Capacitor
Description:
We know that, on storing charges, energy is stored within a capacitor. When a charge is stored in a capacitor, it repels all other incoming charges within capacitor. So in order to store charge, energy is applied to work against the force of repulsion. Theses energy increases with the increase in number of charge in a capacitor.
To know how much work is done to store charge in a capacitor, Take a parallel plate capacitor. Let the charge stored within charge carrying plate be ‘q’and potential due to total charge be ‘V’. If W is the total work done to store charge q.
Method 1
Let Winitial be the initial work done to move first charge to the capacitor, Wfinal be the final work done to move last charges to the capacitor against the force of repulsion and Waverage be the average of total work done.
Initially the charge on the capacitor was zero, so the potential was also zero.
So, Winitial = 0
finally the charge on the capacitor is q, and potential due to it is V
So, Wfinal = qV
So, Waverage = qV + 0/2 = qV/2
Waverage = qV/2 = energy
So, the total energy stored within the capacitor is qV/2.
Method 2
When there was no charge within the capacitor, the potential of it was 0. After a few charge was given to the capacitor, let there is a small rise in potential dV. So the small work done due to small rise in potential will be:
dW = qdV
So total work done when voltage is increased from 0 to V will be:
But, q = CV
C = q/v
Since, C = Kε0A/d
But, K and ε0 are constants
Distance (d) between the plates of capacitor and area(A) of capacitor do not changes during the charging process. So, ‘d’ and ‘A’ are also constant.
C is constant
W = CV2/2 = 1/2 CV2
This work is converted to energy so,
U = 1/2 CV2 ------(1)
But CV = q
So, W = 1/2 qV
This work is converted to energy so,
U = 1/2 qV ------(2)
Energy can also be converted in terms of C and q.
As CV = q
V = q/C
So, U = 1/2c q2 ------(3)
Equation (1), (2), (3) gives the formula for potential stored within a capacitor.