Question 156: Electrostats & Capacitance

Question

The distance between two plates of a capacitor is $$\mathrm{d}$$ and its capacitance is $$\mathrm{C}_{1}$$, when air is the medium between the plates. If a metal sheet of thickness $$\frac{2 d}{3}$$ and of the same area as plate is introduced between the plates, the capacitance of the capacitor becomes $$\mathrm{C}_{2}$$. The ratio $$\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$$ is

Accepted Answer

When a metal sheet of thickness ($\frac{2d}{3}$) is introduced between the plates of a capacitor, it divides the capacitor into two separate capacitors. The metal sheet acts as a new plate in each capacitor, and because the metal is a conductor, it is at the same potential as the plates on either side.
The capacitance of the original capacitor with air between the plates is given by:
$\mathrm{C}_1 = \frac{\epsilon_0 \mathrm{A}}{\mathrm{d}}$
where ($\epsilon_0$) is the permittivity of free space, ($\mathrm{A}$) is the area of one of the plates, and ($\mathrm{d}$) is the distance between the plates.
When the metal sheet of thickness ($\frac{2d}{3}$) is introduced, the distance between the plates is reduced by ($\frac{2d}{3}$), so the capacitance of each of the new capacitors is:
$\mathrm{C}' = \frac{\epsilon_0 \mathrm{A}}{\mathrm{d}-\frac{2d}{3}}$
Simplifying, we get:
$\mathrm{C}' = \frac{3\epsilon_0 \mathrm{A}}{\mathrm{d}}$
Since the two capacitors are in parallel with each other, the total capacitance is:
$\mathrm{C}_2 = 2\mathrm{C}' = 2 	imes \frac{3\epsilon_0 \mathrm{A}}{\mathrm{d}} = 3\mathrm{C}_1$
So the ratio of the capacitances is:
$\frac{\mathrm{C}_2}{\mathrm{C}_1} = 3:1$

Answer

1 : 1

Answer

3 : 1

Answer

2 : 1

Answer

4 : 1

Electrostats & Capacitance

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