Electrostatics 5 Question 39
41. A parallel plate capacitor of plate area $A$ and plate separation $d$ is charged to potential difference $V$ and then the battery is disconnected. A slab of dielecric constant $K$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $Q, E$ and $W$ denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted), and work done on the system, in question, in the process of inserting the slab, then
(a) $Q=\frac{\varepsilon _0 A V}{d}$
(b) $Q=\frac{\varepsilon _0 K A V}{d}$ (1991, 2M)
(c) $E=V / K d$
(d) $W=\frac{\varepsilon _0 A V^{2}}{2 d}[1-1 / K]$
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Solution:
- Battery is removed. Therefore, charge stored in the plates will remain constant.
$$ Q=C V=\frac{\varepsilon _0 A}{d} V \text { or } Q=\text { constant. } $$
Now, dielectric slab is inserted. Therefore, $C$ will increase. New capacity will be,
$$ C^{\prime}=K C=\frac{\varepsilon _0 K A}{d} \Rightarrow V^{\prime}=\frac{Q}{C^{\prime}}=\frac{V}{K} $$
and new electric field, $E=\frac{V^{\prime}}{d}=\frac{V}{K \cdot d}$
Potential energy stored in the capacitor,
Initially, $\quad U _i=\frac{1}{2} C V^{2}=\frac{\varepsilon _0 A V^{2}}{2 d}$
Finally, $U _f=\frac{1}{2} C^{\prime} V^{\prime 2}=\frac{1}{2} \frac{K \varepsilon _0 A}{d} \quad \frac{V^{2}}{K}=\frac{\varepsilon _0 A V^{2}}{2 K d}$
Work done on the system will be
$$ |\Delta U|=\frac{\varepsilon _0 A V^{2}}{2 d} 1-\frac{1}{K} $$
$\therefore$ Correct options are (a), (c) and (d).
