Answer:
Electric field, E = 45.19 N/C
Explanation:
It is given that,
Surface charge density of first surface, [tex]\sigma_1=0.2\ nC/m^2=0.2\times 10^{-9}\ C/m^2[/tex]
Surface charge density of second surface, [tex]\sigma=-0.6\ nC/m^2=-0.6\times 10^{-9}\ C/m^2[/tex]
The electric field at a point between the two surfaces is given by :
[tex]E=\dfrac{\sigma}{2\epsilon_o}[/tex]
[tex]E=\dfrac{\sigma1-\sigma_2}{2\epsilon_o}[/tex]
[tex]E=\dfrac{0.2\times 10^{-9}-(-0.6\times 10^{-9})}{2\times 8.85\times 10^{-12}}[/tex]
E = 45.19 N/C
So, the magnitude of the electric field at a point between the two surfaces is 45.19 N/C. Hence, this is the required solution.
