Question

A variable capacitor is kept connected to a $10V$ battery . If the capacitance of the capacitor is changed from $7\mu$ F to $3\mu$ F, the change in energy of the capacitor is:

• A. $2\times {10}^{-4}J$
• B. $4\times {10}^{-4}J$
• C. $6\times {10}^{-4}J$
• D. $8\times {10}^{-4}J$

Answer 1

The correct option is A $2\times {10}^{-4}J$

We know energy $E=\frac{1}{2}c{v}^2$
$\Rightarrow E_1=\frac{1}{2}(7\times {10}^{-6})\times (10{)}^2$
$\Rightarrow E_1=3.5\times (10{)}^{-4}$
$\Rightarrow E_2=\frac{1}{2}(3\times {10}^{-6})\left({10}^2\right)$
$\Rightarrow E_2=1.5\times (10{)}^{-4}$
Change in energy$=E_1-E_2$
$=(3.5-1.5)\times {10}^{-4}$
$=2\times {10}^{-4}J$