Question

For a parallel plate capacitor prove that the total energy stored in a capacitor is 1/2 cv2 and here derive expression for the energy density of the capacitor

Answer 1

Answer:

Explanation:

Refer to pictures

Answer 2

Answer:

Expression for the energy density of the capacitor is  $\frac{1}{2}\epsilon_{o}}E^2$ where ε₀ is absolute permittivity of free space and E is electric field.

Explanation:

Consider that C is the capacitance of a capacitor. Initially the capacitor was not charged then connected to battery and charged to potential V and 'q' is the charge of parallel plate capacitor,

$q = CV$

Consider that small charge 'dq' to capacitor at constant potential V. The small work done will be:

dW = Vdq = (q/C)dq

The total amount of work done is;

$W=\int\limits^q_0 {\frac{q}{C} } \, dq=\frac{1}{C}\int\limits^q_0 {q} \, dq$

$W=\frac{1}{2} \frac{q^2}{C}$                                                                         ................(1)

The charge of capacitor is given by $q=CV$ put in  equation (1):

$W=\frac{1}{2}\frac{(CV)^2}{C}$

$W=U=\frac{1}{2} CV^2$                                                              ...............(2)

The above expression gives the total energy stored in a capacitor.

Expression for energy density of the capacitor:

The electric field inside the capacitor will be given by:

$E=\frac{\sigma}{\epsilon_{o}}=\frac{q}{A\epsilon_{o}}$                                                                 .................(3)

Capacitance is given by: $C=\frac{\epsilon_{o}A}{d}$

$\epsilon_{o}A= Cd$ Put the is value in equation (3);

q = ECd

Substitute the value of 'q' in the equation (1), we get;

$W=U=\frac{E^2C^2d^2}{2C}$

$U=\frac{1}{2}E^2\epsilon_{o}Ad$

Energy density of the capacitor $=\frac{U}{Ad}=\frac{1}{2}\epsilon_{o}}E^2$