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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 $Q_{0}, V_{0}, E_{0}, U_{0}$ respectively. A dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities now given by Q, V, E and U are related with previous ones as :

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$V > V_{0}$

$E > E_{0}$

$Q > Q_{0}$

$U > U_{0}$

answer is C, D.

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Detailed Solution

Where dielectric slab is introduced capacity gets increases while potential difference remains unchanged
$V = V_{0} C > C_{0} Q = CV Q > Q_{0} V = \frac{1}{2} {CV}^{2} \Rightarrow U > U_{0}$

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