Question: Q. 26. (i) Deduce the expression for the electrostatic energy stored in a capacitor of capacitance ’ $C$ ’ and having charge ’ $Q$ ‘.
(ii) How will the (a) energy stored and (b) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant ’ $\mathrm{K}$ ’ ?
U] [O.D. I, II, III 2012]
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Solution:
Ans. (i) Expression of electrostatic energy stored in a capacitor
(ii) Effect on energy stored and electric field in capacitor due to dielectric material. Potential difference between the plates of capacitor
$$ V=q / C $$
Work done in adding an additional charge $d q$ on the capacitor
$$ \begin{align*} d W & =V \times d q \ & =\left(\frac{q}{C}\right) \times d q \end{align*} $$
$\therefore$ Total energy stored in the capacitor,
$$ \begin{equation*} U=\int d \mathrm{~W}=\int_{0}^{Q} \frac{q}{C} d q=\frac{1}{2} \frac{Q^{2}}{C} \tag{1} \end{equation*} $$
When battery is disconnected:
(a) Energy stored will be decreased or energy stored $=\frac{1}{\kappa}$ times the initial energy.
$1 / 2$
(b) Electric field would decrease.
[CBSE Marking Scheme 2012]
