Electrostatics 5 Question 41
43. 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 the battery still in connection. The corresponding quantities now given by $Q, V, E$ and $U$ are related to the previous one as
(1985, 2M)
(a) $Q>Q _0$
(c) $E>E _0$
(b) $V>V _0$
(d) $U>U _0$
Numerical Value Based Question
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Solution:
- When dielectric slab is introduced capacity gets increased while potential difference remains unchanged.
$$ \begin{aligned} \therefore \quad V & =V _0, C>C _0 \\ Q & =C V \quad \therefore \quad \quad Q>Q _0 \\ U & =\frac{1}{2} C V^{2} \quad \therefore \quad U>U _0 \\ E & =\frac{V}{d} \text { but } V \text { and } d \text { both are unchanged. } \end{aligned} $$
Therefore, $\quad E=E _0$
Therefore, correct options are (a) and (d).
