- Home
- Standard 12
- Physics
2. Electric Potential and Capacitance
medium
A parallel plate capacitor has plates with area $A$ and separation $d$ . A battery charges the plates to a potential difference $V_0$ . The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced, is
A
$K$
B
$1/K$
C
$\frac{A}{{{d^2}K}}$
D
$\frac{{{d^2}K}}{A}$
Solution
$\mathrm{U}=\mathrm{Q}^{2} / 2 \mathrm{C}$
Now, $\mathrm{C}^{\prime}=\mathrm{KC}$
As battery is disconnected, $\mathrm{Q}$ remains unaltered, so,
$\mathrm{U}^{\prime}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}^{\prime}}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{KC}}$
$\therefore \frac{\mathrm{U}}{\mathrm{U}^{\prime}}=\mathrm{K}$
Standard 12
Physics
