Answer:
[tex]\sigma=3.72\times 10^{-6}\ C/m^2[/tex]
Explanation:
Given that,
Charge, [tex]q=1.4\ \mu C=1.4\times 10^{-6}\ C[/tex]
Force experienced by the charge, F = 0.59 N
We need to find the magnitude of charge density on either plate of the capacitor. Electric field on parallel plate capacitor is given by :
[tex]E=\dfrac{\sigma}{\epsilon_o}[/tex]
[tex]E=\dfrac{F}{q}[/tex]
[tex]\dfrac{F}{q}=\dfrac{\sigma}{\epsilon_o}[/tex]
[tex]\sigma=\dfrac{F.\epsilon_o}{q}[/tex]
[tex]\sigma=\dfrac{0.59\times 8.85\times 10^{-12}}{1.4\times 10^{-6}}[/tex]
[tex]\sigma=3.72\times 10^{-6}\ C/m^2[/tex]
So, the charge density on either plate of the capacitor is [tex]3.72\times 10^{-6}\ C/m^2[/tex]. Hence, this the required solution.
