Question 65: Electrostatics & Capacitance

Question

In the figure, first the capacitors are fully charged by closing the key $$\mathrm{K}$$. Then after opening the Key a dielectric material with dielectric constant 2 is filled in the space between the plates of both the capacitor. At this state the ratio of the Charge on the capacitor $$C_1$$ to that of $$C_2$$ is:

Accepted Answer

The capacitance of a conductor is defined as the ratio of the charge given to the rise in the potential of the conductor.
$$\begin{aligned}
& C=\frac{q}{V}=\frac{q^2}{W} \quad\left(V=\frac{W}{q}ight) \
& C=\frac{	ext { ampere }^2-	ext { second }^2}{\mathrm{~kg}^2-	ext { metre }^2 	ext { second }^{-2}}
\end{aligned}$$
Hence, dimensionally $$\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2ight]$$
From Ohm's law, $$V=I R$$,
Hence,
$$\begin{aligned}
R & =\frac{V}{I}=\frac{	ext { Volt }}{	ext { Ampere }}=\frac{W}{q} 	imes \frac{t}{q} \
R & =\frac{F 	imes s 	imes t}{q^2}=\mathrm{kg} \frac{\mathrm{m}}{\mathrm{s}^2} 	imes \frac{\mathrm{m} 	imes \mathrm{s}}{\mathrm{A}^2 	imes \mathrm{s}^2} \
& =\mathrm{kg} \mathrm{m}^2 \mathrm{~s}^{-3} \mathrm{~A}^{-2}
\end{aligned}$$
Dimensionally, $$R=\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}ight]$$
Hence,
$$\begin{aligned}
C R & =\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2ight]\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}ight] \
& =\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}ight]
\end{aligned}$$

Answer

1 : 1

Answer

3 : 2

Answer

2 : 1

Answer

1 : 2

Electrostatics & Capacitance

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