Question 116: Electrostats & Capacitance

Question

A parallel plate capacitor with plate separation 5 mm is charged by a battery. On introducing a mica sheet of 2 mm and maintaining the connections of the plates with the terminals of the battery, it is found that it draws $25 \%$ more charge from the battery. The dielectric constant of mica is $\_\_\_\_$

Accepted Answer

Initially, the capacitor has air between the plates with separation $\mathrm{d}=5 \mathrm{~mm}=5 	imes 10^{-3} \mathrm{~m}$.
The initial capacitance ( $\mathrm{C}_0$ ) is:
$$ C_0=\frac{\epsilon_0 A}{d}=\frac{\epsilon_0 A}{5} $$
Where A is the area of the plates and d is in mm ).
A mica sheet of thickness $\mathrm{t}=2 \mathrm{~mm}$ and dielectric constant K is introduced. The new capacitance $\left(\mathrm{C}^{\prime}ight)$ of a capacitor partially filled with a dielectric is given by:
$$ C^{\prime}=\frac{\epsilon_0 A}{d-t+\frac{t}{K}} $$
Substituting the values $\mathrm{d}=5$ and $\mathrm{t}=2$ :
$$ C^{\prime}=\frac{\epsilon_0 A}{5-2+\frac{2}{K}}=\frac{\epsilon_0 A}{3+\frac{2}{K}} $$
The battery remains connected, meaning the potential difference (V) stays constant.
Initial charge: $\mathrm{Q}_0=\mathrm{C}_0 \mathrm{~V}$
Final charge: $Q^{\prime}=C^{\prime} V$
The capacitor draws $25 \%$ more charge :
$$ \mathrm{Q}^{\prime}=\mathrm{Q}_0+0.25 \mathrm{Q}_0=1.25 \mathrm{Q}_0 $$
$$\Rightarrow $$ $\mathrm{C}^{\prime} \mathrm{V}=1.25 \mathrm{C}_0 \mathrm{~V}$
$\Rightarrow C^{\prime}=\frac{5}{4} C_0$
$$\Rightarrow $$ $\frac{\epsilon_0 \mathrm{~A}}{3+\frac{2}{\mathrm{~K}}}=\frac{5}{4}\left(\frac{\epsilon_0 \mathrm{~A}}{5}ight)$
$$\Rightarrow $$ $\frac{1}{3+\frac{2}{K}}=\frac{1}{4}$
$$\Rightarrow $$ $3+\frac{2}{K}=4$
$$\Rightarrow $$ $\frac{2}{\mathrm{~K}}=1$
$$\Rightarrow $$ $$ K=2 $$
The dielectric constant K is 2.0 .
Hence, the correct option is (D).

Answer

1.0

Answer

2.5

Answer

1.5

Answer

2.0

Electrostats & Capacitance

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