Electrostatics Ques 140
- The figure shows two identical parallel plate capacitors connected to a battery with the switch $S$ closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant (or relative permittivity) 3. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.
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Answer:
Correct Answer: 140.$\frac{3}{5}$
Solution:
Formula:
Distribution of Charges on Connecting Two Charged Capacitors:
- Before opening the switch potential difference across both the capacitors is $V$, as they are in parallel. Hence, energy stored in them is,
$$ \begin{aligned} U_{A} & =U_{B}=\frac{1}{2} C V^{2} \\ \therefore \quad U_{\text {Total }} & =C V^{2}=U_{i} \cdots(i) \end{aligned} $$
After opening the switch, potential difference across it is $V$ and its capacity is $3 C$
$\therefore \quad U_{A}=\frac{1}{2}(3 C) V^{2}=\frac{3}{2} C V^{2}$
In case of capacitor $B$, charge stored in it is $q=C V$ and its capacity is also $3 C$.
Therefore, $U_{B}=\frac{q^{2}}{2(3 C)}=\frac{C V^{2}}{6}$
$\therefore \quad U_{\text {Total }}=\frac{3 C V^{2}}{2}+\frac{C V^{2}}{6}=\frac{10}{6} C V^{2}=\frac{5 C V^{2}}{3}=U_{f} \cdots(ii)$
From Eqs. (i) and (ii), we get $\frac{U_{i}}{U_{f}}=\frac{3}{5}$
BETA
