A parallel plate capacitor has plates with area A and separation d . A battery charges plates to a p

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1. Electric Charges and Fields

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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

$\frac {1}{K}$

C

$\frac {A}{d^2K}$

D

$\frac {d^2K}{A}$

Solution

$\mathrm{U}=\mathrm{Q}^{2} / 2 \mathrm{C}$

Now, $\quad \mathrm{C}^{\prime}=\mathrm{KC}$

As battery is disconnected, $\mathrm{Q}$ remains unaltered.

So, $U^{\prime}=\frac{Q^{2}}{2 C^{\prime}}=\frac{1}{2} \frac{Q^{2}}{K C}$

$\therefore \frac{\mathrm{U}}{\mathrm{U}^{\prime}}=\frac{\mathrm{Q}^{2} / 2 \mathrm{C}}{\mathrm{Q}^{2} / 2 \mathrm{KC}}=\mathrm{K}$

Standard 12

Physics

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