Question

A parallel-plate capacitor has a plate area of 0.3 m2 and a plate separation of 0.1 mm. If the charge on each plate has a magnitude of 5 × 10–6 C then the force exerted by one plate on the other has a magnitude of about:

Answer 1

Answer:

4.72 N

Explanation:

The charge density across each plate is given by:

$\sigma = \frac{Q}{A}$

where

$Q=5\cdot 10^{-6}C$ is the charge on each plate

$A=0.3 m^2$ is the area of each plate

Solving,

$sigma = \frac{5\cdot 10^{-6}C}{0.3 m^2}=1.67\cdot 10^{-5} C/m^2$

The force exerted by one plate on the other is given by:

$F=\frac{Q\sigma}{2\epsilon_0}$

where

$Q=5\cdot 10^{-6}C$ is the charge on each plate

$\sigma=1.67\cdot 10^{-5} C/m^2$ is the surface charge density

$\epsilon_0$ is the vacuum permittivity

Substituting,

$F=\frac{(5\cdot 10^{-6} C)(1.67\cdot 10^{-5} C/m^2)}{2(8.85\cdot 10^{-12}F/m)}=4.72 N$