An air capacitor is connected to a battery of emf V. A dielectric medium of dielectric constant K is introduced between the plates of this capacitor, with the battery still connected. The final charge on the capacitor plates relative to the previous charge

is less

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is more

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is the same

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may be more or less depending upon the dielectric constant of the medium

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Solution

The correct option is B is more
REF.Image
Initially $Q = C V$
$= \frac{A ε_{0}}{d} V$
$(k = 1)$
But when dielectric
is inserted
$Q^{1} = K \frac{A ε_{0}}{d} V$
$k > 1$
so
$Q^{1} > Q$
(B) option is correct.

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An air capacitor is connected to a battery of emf V. A dielectric medium of dielectric constant K is introduced between the plates of this capacitor, with the battery still connected. The final charge on the capacitor plates relative to the previous charge

The correct option is B is moreREF.ImageInitially Q=CV=Aε0dV(k=1)But when dielectricis insertedQ1=KAε0dVk>1soQ1>Q(B) option is correct.

The correct option is B is more
REF.Image
Initially $Q = C V$
$= \frac{A ε_{0}}{d} V$
$(k = 1)$
But when dielectric
is inserted
$Q^{1} = K \frac{A ε_{0}}{d} V$
$k > 1$
so
$Q^{1} > Q$
(B) option is correct.