derive an expression for the energy stored in a capacitor with Air as the medium between its plates.how does the stored energy change if Air is replaced by a medium of dielectric constant k
Let us consider a capacitor of capacitance C and potential difference V between the plates.
Let the charge on one plate be +q and -q on the other.
Suppose the capacitor is being charged gradually.
Now,at any stage the charge on capacitor is q.
Therefore, the potential difference = q/C
Small amount of work done in giving n additional charge dq to the capacitor is =
dW = C/q × dq
Total work done in giving a charge Q to the capacitor is
W=∫dW ~~~~~~ (A) equation
Now, by putting the value of dW = C/q × dq in above equation (A) and integrating it and putting values of upper limit = Q and lower limit = 0 and thus integrating the whole equation we get,
W = Q^2/ C
Here let Energy= E
E= Q^2/ 2C
= C V^2 / 2
= QV/2
The energy is stored in the form of potential energy.
Hope that helps
