Question 522: Electrostats & Capacitance

Question

Two circular metal plates each of radius ' $r$ ' are kept parallel to each other distance ' $d$ ' apart. The capacitance of the capacitor formed is ' $\mathrm{C}_1$ '. If the radius of each of the plates is increased to $\sqrt{2}$ times the earlier radius and their distance of separation decreased to half the initial value, the capacitance now becomes ' $\mathrm{C}_2$ '. The ratio of the capacitances $\mathrm{C}_1: \mathrm{C}_2$ is

Accepted Answer

The capacitance of a parallel plate capacitor is given by the formula:
$$ C = \frac{\varepsilon_0 A}{d} $$
where $ \varepsilon_0 $ is the permittivity of free space, $ A $ is the area of the plates, and $ d $ is the distance between the plates.
Initial Capacitance $ C_1 $:
The area of each plate, $ A_1 $, is:
$$    A_1 = \pi r^2    $$
So, the initial capacitance $ C_1 $ is:
$$ C_1 = \frac{\varepsilon_0 \pi r^2}{d} $$
Changed Capacitance $ C_2 $:
The radius of each plate is increased to $ \sqrt{2} 	imes r $, so the new area $ A_2 $ is:
$$    A_2 = \pi (\sqrt{2}r)^2 = 2\pi r^2    $$
The distance between the plates is decreased to $ \frac{1}{2}d $.
The new capacitance $ C_2 $ is:
$$ C_2 = \frac{\varepsilon_0 \cdot 2\pi r^2}{\frac{1}{2}d} = \frac{2 \cdot \varepsilon_0 \pi r^2}{\frac{1}{2}d} = \frac{4 \varepsilon_0 \pi r^2}{d} $$
Ratio $ C_1 : C_2 $:
$$ \frac{C_1}{C_2} = \frac{\frac{\varepsilon_0 \pi r^2}{d}}{\frac{4 \varepsilon_0 \pi r^2}{d}} = \frac{1}{4} $$
Therefore, the ratio of the capacitances $ C_1 : C_2 $ is:
Option C: $ 1: 4 $

Answer

$1: 1$

Answer

$1: 2$

Answer

$1: 4$

Answer

$4: 1$

Electrostats & Capacitance

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