Electrostatics Ques 127
- 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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Answer:
Correct Answer: 127.(a,c,d)
Solution:
Formula:
- 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).
BETA
