The energy and capacity of a charged parallel plate capacitor are $$E$$ and $$C$$ respectively. If a dielectric slab of ${}_{}$$$E_r=6$$ is inserted in it, then the energy and capacity become:
(Assuming the charge on plates remains constant)

 1 $$6 \mathrm E,~6 \mathrm C$$ 2 $$\mathrm E,~ \mathrm C$$ 3 $${E \over 6},~6 \mathrm C$$ 4 $$\mathrm E,~6 \mathrm C$$
Subtopic: Â Energy stored in Capacitor |
Â 75%
From NCERT
AIPMT - 1999
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Energy per unit volume for a capacitor having area $$A$$ and separation $$d$$ kept at a potential difference $$V$$ is given by:
1. $$\frac{1}{2}\varepsilon_0\frac{V^2}{d^2}$$
2. $$\frac{1}{2}\frac{V^2}{\varepsilon_0d^2}$$
3. $$\frac{1}{2}CV^2$$
4. $$\frac{Q^2}{2C}$$

Subtopic: Â Energy stored in Capacitor |
Â 81%
From NCERT
AIPMT - 2001
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