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A parallel-plate capacitor of plate area 10 cm2 and plate
separation 3 mm is charged to a potential difference 12 V
and then the battery is disconnected. A slab of dielectric
constant 3 is then inserted between the plates. The work
done on the system in the process of inserting the slab is
$\alpha \varepsilon_0$. The value of a is
(Take $\varepsilon_0$as the permittivity of free space)
Options:
separation 3 mm is charged to a potential difference 12 V
and then the battery is disconnected. A slab of dielectric
constant 3 is then inserted between the plates. The work
done on the system in the process of inserting the slab is
$\alpha \varepsilon_0$. The value of a is
(Take $\varepsilon_0$as the permittivity of free space)
Solution:
1738 Upvotes
Verified Answer
The correct answer is:
16
Given that Area (A) $=10 \times 10^{-4} \mathrm{~m}^2$
Separation (d) $=3 \times 10^{-3} \mathrm{~m}$
Potential difference $=12 \mathrm{~V}$
Dielectric constant $\mathrm{K}=3$
Work done on the system is a change in potential energy.
So work done will be
$$
\begin{aligned}
& |\mathrm{W}|=\mathrm{V}_{\mathrm{f}}-\mathrm{V}_{\mathrm{j}} \\
& =\frac{\mathrm{Q}^2}{2 \mathrm{KC}}-\frac{\mathrm{Q}^2}{2 \mathrm{C}} \quad \because \mathrm{Q}=\mathrm{CV}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \mathrm{~V} \\
& =\frac{\mathrm{C}^2 \mathrm{~V}^2}{2 \mathrm{KC}}-\frac{\mathrm{C}^2 \mathrm{~V}^2}{2 \mathrm{C}}=\frac{\varepsilon_0 \mathrm{AV}^2}{2 \mathrm{~d}}\left(\frac{1}{\mathrm{~K}}-1\right) \\
& \frac{\varepsilon_0 \times 10 \times 10^{-4} \times 12^2}{2 \times 3 \times 10^{-3}}\left(\frac{1}{3}-1\right)=16 \varepsilon_0
\end{aligned}
$$
Separation (d) $=3 \times 10^{-3} \mathrm{~m}$
Potential difference $=12 \mathrm{~V}$
Dielectric constant $\mathrm{K}=3$
Work done on the system is a change in potential energy.
So work done will be
$$
\begin{aligned}
& |\mathrm{W}|=\mathrm{V}_{\mathrm{f}}-\mathrm{V}_{\mathrm{j}} \\
& =\frac{\mathrm{Q}^2}{2 \mathrm{KC}}-\frac{\mathrm{Q}^2}{2 \mathrm{C}} \quad \because \mathrm{Q}=\mathrm{CV}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \mathrm{~V} \\
& =\frac{\mathrm{C}^2 \mathrm{~V}^2}{2 \mathrm{KC}}-\frac{\mathrm{C}^2 \mathrm{~V}^2}{2 \mathrm{C}}=\frac{\varepsilon_0 \mathrm{AV}^2}{2 \mathrm{~d}}\left(\frac{1}{\mathrm{~K}}-1\right) \\
& \frac{\varepsilon_0 \times 10 \times 10^{-4} \times 12^2}{2 \times 3 \times 10^{-3}}\left(\frac{1}{3}-1\right)=16 \varepsilon_0
\end{aligned}
$$
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