electrostatic-potential-and-capacitance Question 35
Question: Q. 5. A parallel plate capacitor of capacitance $C$ is charged to a potential $V$ by a battery. Without disconnecting the battery, the distance between the plates is tripled and a dielectric medium of $\kappa=10$ is introduced between the plates of the capacitor. Explain giving reasons, how will the following be affected :
(i) capacitance of the capacitor
(ii) charge on the capacitor, and
(iii) energy density of the capacitor.
U] [O.D. (Compt) I, II, III 2017]
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Solution:
Ans. Effect on capacitance
Effect on charge
Effect on energy
(i)
$$ \begin{aligned} C & =\frac{\varepsilon_{0} A}{d} \ C^{\prime} & =\frac{\kappa \varepsilon_{0} A}{d^{\prime}}=10 \frac{\varepsilon_{0} A}{3 d}=\frac{1 / 2}{3} e_{1 / 2}^{1} \end{aligned} $$
(ii) $V$ remains same since battery is not disconnected
$$ \therefore \quad Q^{\prime}=E^{\prime} V=\frac{10}{3} C V=\frac{10}{3} Q $$
(iii) Energy density, $4{ }{T}=\frac{1}{2} \varepsilon{0} E^{2}$
$$ \begin{align*} E & =\frac{(V)}{d} \ U_{d}^{\prime} & =\frac{1}{2} \kappa \varepsilon_{0} E^{\prime 2} \ & =\frac{10}{2} \varepsilon_{0}\left(\frac{V}{d^{\prime}}\right)^{2} \ & =\frac{10}{9}\left(\frac{1}{2} \varepsilon_{0} E^{2}\right) \ & =\frac{10}{9} U_{d} \end{align*} $$
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