Respuesta :
Explanation:
Relation between electric field and charge density is as follows.
E = [tex]\frac{\sigma}{2 \epsilon}[/tex]
where, [tex]\sigma[/tex] = charge density
[tex]\epsilon[/tex] = permittivity of free space = [tex]8.85 \times 10^{-12}[/tex]
So, [tex]E_{\text{outside}}[/tex] = 0
[tex]E_{inside} = \frac{+\sigma}{2 \epsilon} - \frac{-\sigma}{2 \epsilon}[/tex]
or, [tex]E_{inside} = \frac{\sigma}{\epsilon}[/tex]
Now, formula to calculate the potential difference of two conductors is as follows.
[tex]V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}[/tex]
It is given that,
d = 6.0 mm = [tex]6 \times 10^{-3} m[/tex]
[tex]\sigma = 40 \times 10^{-12} C/m^{2}[/tex]
Hence, we will calculate the magnitude of the electric potential differences between the two conductors as follows.
[tex]V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}[/tex]
= [tex]\frac{40 \times 10^{-12} \times 6 \times 10^{-3}}{8.85 \times 10^{-12}}[/tex]
= 0.0271 volts
thus, we can conclude that value of the magnitude of the electric potential differences between the two conductors is 0.0271 volts.
