Question

Two flat conductors are placed with their inner faces separated by 6.0 mm. If the surface charge density on one of the inner faces is 40 pC/m2, what is the magnitude of the electric potential differences between the two conductors?

Answer 1

Explanation:

Relation between electric field and charge density is as follows.

           E = $\frac{\sigma}{2 \epsilon}$

where,    $\sigma$ = charge density

              $\epsilon$ = permittivity of free space = $8.85 \times 10^{-12}$

So,  $E_{\text{outside}}$ = 0

      $E_{inside} = \frac{+\sigma}{2 \epsilon} - \frac{-\sigma}{2 \epsilon}$

or,     $E_{inside} = \frac{\sigma}{\epsilon}$

Now, formula to calculate the potential difference of two conductors is as follows.

         $V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}$

It is given that,

           d = 6.0 mm = $6 \times 10^{-3} m$

        $\sigma = 40 \times 10^{-12} C/m^{2}$

Hence, we will calculate the magnitude of the electric potential differences between the two conductors as follows.

        $V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}$

                     = $\frac{40 \times 10^{-12} \times 6 \times 10^{-3}}{8.85 \times 10^{-12}}$      

                     = 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.