Electrostatics 5 Question 50
52. 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
Show Answer
Solution:
- Before opening the switch potential difference across both the capacitors is $V$, as they are in parallel. Hence, energy stored in them is,
$$ \begin{aligned} U _A & =U _B=\frac{1}{2} C V^{2} \\ \therefore \quad U _{\text {Total }} & =C V^{2}=U _i \end{aligned} $$
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, $U _B=\frac{q^{2}}{2(3 C)}=\frac{C V^{2}}{6}$
$\therefore \quad U _{\text {Total }}=\frac{3 C V^{2}}{2}+\frac{C V^{2}}{6}=\frac{10}{6} C V^{2}=\frac{5 C V^{2}}{3}=U _f$
From Eqs. (i) and (ii), we get $\frac{U _i}{U _f}=\frac{3}{5}$
