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

0.07 V

Step-by-step explanation:

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

`E=(\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=(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`