Electrostatics 5 Question 13
14. A parallel plate capacitor having capacitance $12 \mathrm{pF}$ is charged by a battery to a potential difference of $10 \mathrm{~V}$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is
(a) $560 \mathrm{pJ}$
(b) $508 \mathrm{pJ}$
(c) $692 \mathrm{pJ}$
(d) $600 \mathrm{pJ}$
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Answer:
Correct Answer: 14. (b)
Solution:
- Energy stored in a charged capacitor is given by
$ U=\frac{1}{2} C V^{2}=\frac{1}{2} \cdot \frac{Q^{2}}{C} $
Here, $C=12 \times 10^{-12} \mathrm{~F}$ and $V=10 \mathrm{~V}$.
$ \begin{aligned} \Rightarrow \quad U & =\frac{1}{2} \times 12 \times 10^{-12} \times 100 \\ U & =6 \times 10^{-10} \mathrm{~J} \end{aligned} $
After insertion of slab, capacitance will be
$ C^{\prime}=K C \text { and final energy, } $
$ \begin{aligned} U^{\prime} & =\frac{1}{2} \cdot \frac{Q^{2}}{C^{\prime}}=\frac{1}{2} \frac{Q^{2}}{K C} \\ \Rightarrow U^{\prime} & =\frac{1}{K} U=\frac{1}{6.5} \times 6 \times 10^{-10} \mathrm{~J} \end{aligned} $
So, energy dissipated in the process will be equal to work done on the slab, i.e.
$ \begin{aligned} \Delta U & =U-U^{\prime}=1-\frac{1}{6.5} \times 6 \times 10^{-10} \mathrm{~J} \\ \Rightarrow \Delta U & =\frac{5.5}{6.5} \times 6 \times 10^{-10} \mathrm{~J} \\ & \cong 5.08 \times 10^{-10} \mathrm{~J} \text { or } 508 \mathrm{pJ} \end{aligned} $
