Question

An infinite plane of charge has surface charge density 7.2 μC/m^2. How far apart are the equipotential surfaces whose potentials differ by 100 V?

Answer 1

so the distance between two points are

`d = 0.246 * 10^(-3) m`

Step-by-step explanation:

Surface charge density of the charged plane is given as

`\sigma = 7.2 \mu C/m^2`

now we have electric field due to charged planed is given as

`E = (\sigma)/(2\epsilon_0)`

now we have

`E = (7.2 * 10^(-6))/(2(8.85 * 10^(-12)))`

`E = 4.07 * 10^5 N/C`

now for the potential difference of 100 Volts we can have the relation as

`E.d = \Delta V`

`4.07 * 10^5 (d) = 100`

`d = (100)/(4.07 * 10^5)`

`d = 0.246 * 10^(-3) m`