Electrostatics Ques 123
- 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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Answer:
Correct Answer: 123.(a,d)
Solution:
Formula:
$$ \begin{aligned} C & =C_{1}+C_{2} \\ C_{1} & =\frac{K \varepsilon_{0} A / 3}{d} \\ C_{2} & =\frac{\varepsilon_{0} 2 A / 3}{d} \\ \Rightarrow \quad C & =\frac{(K+2) \varepsilon_{0} A}{3 d} \\ \Rightarrow \quad \quad \quad \frac{C}{C_{1}} & =\frac{K+2}{K} \end{aligned} $$
Also, $E_{1}=E_{2}=V / d$, where $V$ is potential difference between the plates.
BETA
