A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by Q_o, V₀, E_o and U_o respectively, A dielectric slab is now introduced to fill the space between the plates with battery still in connection. The corresponding quantities now given by Q, V, E and U are related to the previous ones as

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$V > V_{0}$

$E > E_{0}$

$U < U_{0}$

$Q > Q_{0}$

answer is A.

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Detailed Solution

In the presence of dielectric, the capacity $C = {KC}_{0} or C > C_{0}$
We know that $Q = CV \propto C (∵ V = constant)$
$∴ Q > Q_{0}$
Further $U = \frac{1}{2} {CV}^{2} . . . (1)$
$U_{0} = \frac{1}{2} C_{0} V^{2}$
$∴ U > U_{0} (∵ C > C_{0}) . . . . (2)$

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