Question

A condenser of $1\mu F$ is charged to a potential of $1000V$. If a dielectric slab of dielectric constant $5$ is introduced between the plates of the condenser after disconnecting the battery, the loss in the energy of the condenser is :

• A. 0.1 J
• B. 2.5 J
• C. 0.4 J
• D. 5 J

Answer 1

The correct option is C 0.4 J

After disconnecting battery charge present on capacitor becomes constant
$Q=1\times {10}^{-6}\times 1000$
$Q={10}^{-3}C$
When dielectric is added C becomes KC,
$\therefore$ V decreases to $\frac{V}{K}$
$\therefore$ loss in energy $=\frac{1}{2}C{V}^2-\frac{1}{2}\left(KC\right)(\frac{V}{K}{)}^2$
$=\frac{1}{2}C{V}^2-\frac{1}{2}KC\frac{V^2}{K^2}$
$=\frac{1}{2}C{V}^2(1-\frac{1}{K})$
$=\frac{1}{2}\times {10}^{-6}\times {10}^6(1-\frac{1}{5})$
$=\frac{1}{2}\left(\frac{4}{5}\right)$
$=0.4J$