A parallel plate capacitor with air between the plate has a capacitance of 15pF. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant 3.5. Then the capacitance becomes $\frac{x}{4}$ pF. The value of $x$ is ____________.
Answer (integer)
105
Solution
<p>Initially</p>
<p>$\frac{\varepsilon_{0} A}{d}=15 \times 10^{-12} \mathrm{~F}$</p>
<p>Finally</p>
$$
\begin{aligned}
& \frac{3.5 \varepsilon_{0} A}{2 d}=\frac{x}{4} \times 10^{-12} \mathrm{~F} \\\\
& \therefore \frac{3.5}{2} \times 15=\frac{x}{4} \\\\
& \Rightarrow x=\frac{3.5 \times 15 \times 4}{2}=105
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Potential and Capacitance
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