Question

A parallel plate air capacitor has a capacitance of 5.0 F. (a) What is its capacitance if the separation between the plates is decreased to one-half its original value while keeping all other factors constant? (b) What is its capacitance if the area of the plates is decreased to one-half its origi alue while keeping all other factors constant? (c) What is the capacitance if a dielectric of relative permeability of 2 is inserted between the plates of the original air capacitor?

Answer 1

The capacity of the air capacitor is

$\displaystyle C = \frac{\epsilon \epsilon_0 S}{d}$

a) $d_2=0.5d_1$

$\displaystyle C_2 = \frac{\epsilon \epsilon_0 S}{d_2} = \frac{\epsilon \epsilon_0 S}{0.5d_1} = 2 \frac{\epsilon \epsilon_0 S}{d_1 } = 2C_1 = 10 \; F$

b) $S_2 = 0.5 S_1$

$\displaystyle C_2 = \frac{\epsilon \epsilon_0 S_2}{d_2} = \frac{0.5 \epsilon \epsilon_0 S_1}{d} = 0.5 C_1 = 2.5\; F$

c)$\epsilon_2 =2, \epsilon_1=1, \epsilon_2 = 2 \epsilon_1$

$\displaystyle C_2 = \frac{\epsilon_2 \epsilon_0 S}{d} = \frac{2\epsilon_1 \epsilon_0 S}{d} = 2C_1 = 10 \; F$