Question 586: Electrostatics & Capacitance

Question

A parallel plate air capacitor is charged upto 100 V. A plate 2 mm thick is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by 1.6 mm. The dielectric constant of the thick plate is

Accepted Answer

If d is the initial distance between the plates then capacitance is given by $$\mathrm{C=\frac{kA \in_0}{d}}$$
When a plate of thickness t = 2 mm is inserted between the plates the capacitance,
$$C_1=\frac{k A \in_0}{\left(d-t+\frac{t}{k}ight)}$$
If distance between the plates is increased by $$\mathrm{x}=1.6 \mathrm{~mm}$$
Capacitance becomes, $$C_2=\frac{\mathrm{kA} \in_0}{\left(d+x-t+\frac{t}{k}ight)} \quad$$ But $$C_2=C$$
$$\begin{aligned}
& 	herefore \mathrm{d}+\mathrm{x}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{k}}=\mathrm{d} \
& 	herefore \mathrm{t}-\mathrm{x}=\frac{\mathrm{t}}{\mathrm{k}} \
& 	herefore 2-1.6=\frac{2}{\mathrm{k}} \
& 	herefore 0.4=\frac{2}{\mathrm{k}} \quad 	herefore \mathrm{k}=5\end{aligned}$$

Answer

4

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5

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2

Answer

3

Electrostatics & Capacitance

A parallel plate air capacitor is charged upto 100 V. A plate 2 mm thick is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by 1.6 mm. The dielectric constant of the thick plate is

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