First slide

Capacitance

Question

The given fig. shows two identical parallel plate capacitors connected to a battery with switch S is closed. The switch is now opened and free space between the plates of the capacitors is now filled with dielectric of K = 3. Then, ratio of energy stored in both capacitors before and after introduction of dielectric is ;

Moderate

A

$\frac{5}{3}$
B

$\frac{3}{5}$
C

$\frac{7}{5}$
D

$\frac{5}{7}$

Solution

As switch S is closed;
$V_{A} = V_{B} = V ∴ E_{i} = \frac{1}{2} {CV}^{2} + \frac{1}{2} {CV}^{2} = {CV}^{2}$
After introduction of slab as switch S is now opened, $V_{A} = V and C_{A} = 3 C & C_{B} = 3 C$ and Let ${V_{B}}^{'} =$ new pot. difference across capacitor B. Acc. to law of conservation of charge in case of capacitor B;
$CV = 3 {CV}_{B}^{'}$
$\Rightarrow V_{B}^{'} = \frac{V}{3}$
$∴ E_{f} = \frac{1}{2} C_{A} V^{2} + \frac{1}{2} C_{B} V_{B}^{2} = \frac{5}{3} {CV}^{2}$

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