Question 396: Electrostats & Capacitance

Question

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 ____________.

Accepted Answer

Initially
$$
\frac{\varepsilon_{0} A}{d}=15 	imes 10^{-12} \mathrm{~F}
$$
Finally
$$
\begin{aligned}
& \frac{3.5 \varepsilon_{0} A}{2 d}=\frac{x}{4} 	imes 10^{-12} \mathrm{~F} \\
& 	herefore \frac{3.5}{2} 	imes 15=\frac{x}{4} \\
& \Rightarrow x=\frac{3.5 	imes 15 	imes 4}{2}=105
\end{aligned}
$$

Electrostats & Capacitance

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 ____________.

Choose the correct answer: