Answer:
4.72 N
Explanation:
The charge density across each plate is given by:
[tex]\sigma = \frac{Q}{A}[/tex]
where
[tex]Q=5\cdot 10^{-6}C[/tex] is the charge on each plate
[tex]A=0.3 m^2[/tex] is the area of each plate
Solving,
[tex]sigma = \frac{5\cdot 10^{-6}C}{0.3 m^2}=1.67\cdot 10^{-5} C/m^2[/tex]
The force exerted by one plate on the other is given by:
[tex]F=\frac{Q\sigma}{2\epsilon_0}[/tex]
where
[tex]Q=5\cdot 10^{-6}C[/tex] is the charge on each plate
[tex]\sigma=1.67\cdot 10^{-5} C/m^2[/tex] is the surface charge density
[tex]\epsilon_0[/tex] is the vacuum permittivity
Substituting,
[tex]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[/tex]
