Question 597: Electrostatics & Capacitance

Question

In a parallel plate air capacitor the distance between plates is reduced to one fourth and the space between them is filled with a dielectric medium of constant 2 . If the initial capacity of the capacitor is $4 \mu \mathrm{~F}$. then its new capacity is

Accepted Answer

The capacitance of a parallel plate air capacitor is given by
$$C_0=\frac{\varepsilon_0 A}{d}\quad	ext{.... (i)}$$
where, $\varepsilon_0=$ permittivity of the medium,
$A=$ area of plates
and $\quad d=$ distance between the plates
When the distance between plates is reduced and a dielectric slab is introduced, then the capacitance becomes
$C=\frac{K \varepsilon_0 A}{d_1}\quad	ext{.... (ii)}$
Where, $K=$ dielectric constant of medium.
Here, $K=2, d_1=\frac{d}{4}$ and $C_0=4 \mu \mathrm{~F}=4 	imes 10^{-6} \mathrm{~F}$
From Eq. (ii), we get
$$\begin{aligned}
C & =4\left(\frac{K \varepsilon_0 A}{d_1}ight)=4 K\left(\frac{\varepsilon_0 A}{d}ight) \
& =4 K C_0\quad	ext{.... (iii) [from Eq. (i)]}
\end{aligned}$$
Substituting given values in Eq. (iii), we get
$$C=4 	imes 2 	imes 4=32 \mu \mathrm{~F}$$

Answer

$32 \mu \mathrm{~F}$

Answer

$18 \mu \mathrm{~F}$

Answer

$8 \mu \mathrm{~F}$

Answer

$44 \mu \mathrm{~F}$

Electrostatics & Capacitance

In a parallel plate air capacitor the distance between plates is reduced to one fourth and the space between them is filled with a dielectric medium of constant 2 . If the initial capacity of the capacitor is 4 \mu \mathrm{~F}. then its new capacity is

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