Question

prove that energy stored per unit volume in a capacitor is given by 1/2 Eo E^2 where E is the electric field of the capacitor​

Answer 1

IS the answer is correct  

Answer 2

Answer:

Consider that W is the work done by the source of potential V, in order to store an additional charge dq,

             $W=Vdq$                                                ...................(1)

But we know $V=q/C$ put in eq.(1)

 ∴       $W=\frac{q}{C}dq$                                                  ..................(2)

Therefore, the total work done in storing the charge Q,

          $\int\limits{dW}=\int\limits^Q_0 {\frac{q}{C} } \, dq$

              $W=\frac{1}{2} (\frac{q^{2}}{2} )^{Q}_{0}=\frac{Q^{2}}{2C}$                                 ................(3)

If V is the final potential between the capacitor plates, then $Q=CV$

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

Electrostatic potential energy: $U=\frac{Q^{2}}{2C}=\frac{1}{2} CV^{2}=\frac{1}{2} QV$

Consider that A is the area of the each plate and d is the separation between them. And the space between the plates is filled with a medium has dielectric constant K.

Then capacitance will be: $C=\frac{K\epsilon_{o}A}{d}$

If σ is surface density between the plates then electric field will be:

$E=\frac{\sigma }{K\epsilon_{o}}$

⇒  $\sigma= K\epsilon_{o}E$

Charge on the each plate of the capacitor:

$Q=\sigma A=K\epsilon_{o}EA$

Energy stored by the capacitor:

$U=\frac{Q^{2}}{2C}=\frac{(K\epsilon_{o}EA)^{2}}{2(\frac{K\epsilon_{o}A}{d})} =\frac{1}{2}K\epsilon_{o}E^{2}Ad$

But $Ad=\tau$ is volume of space between capacitor plates.

∴    Energy stored $U=\frac{1}{2}K\epsilon_{o}E^{2}\tau$

The electrostatic energy stored per unit volume,

$U_{\tau}=\frac{U}{\tau}=\frac{1}{2}K\epsilon_{o}E^{2}$

Now the dielectric constant for air, K=1

Therefore, the energy stored in per unit volume of the capacitor:

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

where E is the electric field of the capacitor.

Hence, it is proved.