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 $Q_{0}, V_{0}, E_{0}$ and $U_{0}$ 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 with previous ones as :

V > V_{0}

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U > U_{0}

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Q > Q_{0}

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E > E_{0}

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Solution

The correct options are
B $U > U_{0}$
C $Q > Q_{0}$
As the capacitor is still connect with battery so the potential will remain constant as well as field and the capacitor will be charging up. thus, $V = V_{0}, E = E_{0}, Q > Q_{0}$
As the dielectric insert between the plate so the energy $U_{0} = \frac{1}{2} C V_{0}^{2}$ becomes
$U = \frac{1}{2} k C V_{0}^{2}$ . thus energy will increase. $U > U_{0}$

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Similar questions

Q. A parallel plate 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}

, and

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 to the previous ones are :

A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric filed and energy associated with this capacitor are gives by $Q_{0}, V_{0}, E_{0}$ and $U_{0}$ respectively. A dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities are now given by $Q$ , $V$ , $E$ and $U$ are related to the previous one as: