2. Electric Potential and Capacitance
easy

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in

A

Reduction of charge on the plates and increase of potential difference across the plates

B

Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates

C

Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates

D

None of the above

Solution

(c) Battery in disconnected so $Q$ will be constant as $C \propto \,K$. So with introduction of dielectric slab capacitance will increase using $Q = CV$, $V$ will decrease and using $U = \frac{{{Q^2}}}{{2C}}$, energy will decrease.

Standard 12
Physics

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