Electrostatics 5 Question 35
37. A parallel plate capacitor has a dielectric slab of dielectric constant $K$ between its $Q _1$ plates that covers $1 / 3$ of the area of its plates, as shown in the figure. The total capacitance of the capacitor is $C$ while that of the portion with dielectric in between is $C _1$. When the capacitor is
charged, the plate area covered by the dielectric gets charge $Q _1$ and the rest of the area gets charge $Q _2$. The electric field in the dielectric is $E _1$ and that in the other portion is $E _2$. Choose the correct option/options, ignoring edge effects.
(a) $\frac{E _1}{E _2}=1$
(b) $\frac{E _1}{E _2}=\frac{1}{K}$
(c) $\frac{Q _1}{Q _2}=\frac{3}{K}$
(d) $\frac{C}{C _1}=\frac{2+K}{K}$
(2014 Adv.)
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Solution:
$$ \begin{aligned} C & =C _1+C _2 \\ C _1 & =\frac{K \varepsilon _0 A / 3}{d} \\ C _2 & =\frac{\varepsilon _0 2 A / 3}{d} \\ \Rightarrow \quad C & =\frac{(K+2) \varepsilon _0 A}{3 d} \\ \Rightarrow \quad \quad \quad \frac{C}{C _1} & =\frac{K+2}{K} \end{aligned} $$
Also, $E _1=E _2=V / d$, where $V$ is potential difference between the plates.
