Question

Two flat conductors are placed with their inner faces separated by 9mm. If the surface charge density on inner face A is 69 pC/m^2 and on inner face B is -69 pC/m^2 calculate the electric potential difference using epsilon naught value of 8.85419x10^-12 C^2/Nm^2. Answer in units of V.

Answer 1

Answer:

0.07 V

Explanation:

The electric field between two parallel plates is uniform, and its magnitude is given by

$E=\frac{\sigma}{\epsilon_0}$

where

$\sigma$ is the magnitude of the charge density on each face

$\epsilon_0 = 8.85419 \cdot 10^-12 C^2/Nm^2$ is the vacuum permittivity

In this problem, we have

$\sigma=69 pC/m^2 = 69\cdot 10^{-12} C/m^2$

So, the electric field is

$E=\frac{69\cdot 10^{-12} C/m^2}{8.85419\cdot 10^-12 C^2/Nm^2}=7.79 N/C$

So now we can calculate the potential difference between the two plates, given by:

$\Delta V=E d$

where $d=9 mm=0.009 m$ is the distance between the two plates. Substituting, we find

$\Delta V=(7.79 N/C)(0.009 m)=0.07 V$