A parallel plate capacitor filled with a medium of dielectric constant 10 , is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15 . Then the energy of capacitor will

Question

Aicharya · Accepted Answer

When a dielectric of dielectric constant $K$ is inserted between the plates of a capacitor $C_{0}$ , the capacitance becomes $K C_{0}$ .

Energy stored in a capacitor is given by, $U = \frac{1}{2} C V^{2}$ . Now,

$U_{i} = \frac{1}{2} (K_{1} C_{0}) V^{2}$ and $U_{f} = \frac{1}{2} (K_{2} C_{0}) V^{2}$

$\Rightarrow Δ U = U_{f} - U_{i} = \frac{1}{2} (K_{2} - K_{1}) C_{0} V^{2}$

Percentage change in the energy will be,

$\frac{Δ U}{U_{i}} \times 100 = \frac{\frac{1}{2} \times 5 \times {C_{0}}^{2}}{\frac{1}{2} \times 10 \times {C_{0}}^{2}} \times 100 = 50 %$

A parallel plate capacitor filled with a medium of dielectric constant 10 , is connected across a battery…

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