Question 710: Electrostatics & Capacitance

Question

A medium having dielectric constant $K>1$ fills the space between the plates of a parallel plate capacitor. The plates have large area, and the distance between them is $d$. The capacitor is connected to a battery of voltage $V$, as shown in Figure (a). Now, both the plates are moved by a distance of $\frac{d}{2}$ from their original positions, as shown in Figure (b).
In the process of going from the configuration depicted in Figure (a) to that in Figure (b), which of the following statement(s) is(are) correct?

Accepted Answer

$C_a=\frac{K \varepsilon_0 A}{d}$
$$ 	herefore $$ $q_a=\frac{K \varepsilon_0 A}{d} V$
$ E_a  =\frac{q_a}{K A \varepsilon_0}$
$=\frac{K \varepsilon_0 A V}{d K \varepsilon_0 A} $
$ =\frac{V}{d}$
$\begin{aligned} C_b & =\frac{\varepsilon_0 A}{d+\left(\frac{d}{K}ight)} \\ & =\frac{\varepsilon_0 A K}{d(K+1)}\end{aligned}$
$q_b=\frac{\varepsilon_0 A K V}{d(K+1)}$
$\left(E_bight)_{	ext {dielectric }}  $
$=\frac{E_{	ext {air }}}{K}  $
$=\frac{q_b}{K A \varepsilon_0}  $
$=\frac{\varepsilon_0 A K V}{d(K+1)\left(K A \varepsilon_0ight)}$
$ =\frac{V}{d(K+1)}$
$\begin{aligned} &	herefore	ext {  Capacitance decrease by a factor of } \frac{1}{K+1} \\ & 	ext { Work done in the process }=U_f-U_i \\ &=\frac{1}{2}\left(C_f-C_iight) V^2 \\ &=\frac{1}{2}\left(\frac{\varepsilon_0 A K}{d(K+1)}-\frac{K \varepsilon_0 A}{d}ight) V^2 \\ &=\frac{1}{2} V^2 \frac{\varepsilon_0 A K}{d}\left(\frac{1}{K+1}-1ight) \\ &=\frac{1}{2} \frac{\varepsilon_0 A K V^2}{d} \frac{1-K-1}{K+1} \\ &=\frac{1}{2} \frac{\varepsilon_0 A V^2}{d}\left(\frac{-K^2}{K+1}ight)\end{aligned}$

Answer

The electric field inside the dielectric material is reduced by a factor of $2 K$.

Answer

The capacitance is decreased by a factor of $\frac{1}{K+1}$.

Answer

The voltage between the capacitor plates is increased by a factor of $(K+1)$.

Answer

The work done in the process
DOES NOT
depend on the presence of the dielectric material.

Electrostatics & Capacitance

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