A parallel plate capacitor is connected to a battery. The quantit… - Vidyadip

MCQ

A parallel plate capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with the capacitor are given by $Q_0, V_0, E_0$ and $U_0$ respectively. A dielectric slab is introduced between plates of capacitor but battery is still in connection. The corresponding quantities now given by $Q, V, E$ and $U$ related to previous ones are

✓
$Q > Q_0$
B
$V > V_0$
C
$E > E_0$
D
$U < U_0$

✓

Answer

Correct option: A.

$Q > Q_0$

a
As the battery till in connection so potential will remained unchanged. i,e $V=V_{0}$

$E_{0}=\frac{V_{0}}{d}$ and $E=\frac{V}{d}=\frac{V_{0}}{d} \Rightarrow E=E_{0}$ where $d=$ separation of plates.

Let initial capacitance $=C_{0}$

When dielectric slab with dielectric constant $\mathrm{K}$ is inserted, the capacitance becomes $C=K C_{0}$

Thus, $Q_{0}=C_{0} V_{0}$ and $Q=C V=K C_{0} V_{0}=K Q_{0} \Rightarrow Q>Q_{0}$ as $(K>1)$

$\mathrm{Also}, U_{0}=(1 / 2) C_{0} V_{0}^{2}$ and $U=(1 / 2) C V^{2}=(1 / 2) K C_{0} \cdot V_{0}=K U_{0}$

Hence, $U>U_{0}$

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