An infinite plane of charge has surface charge density 7.2 Î¼C/m^2. How far apart are the equipotential surfaces whose potentials differ by 100 V?

Question

asked Jun 28, 2020 25.7k views

1 Answer

ScubaManDan · Answer 1 · 2020-07-04T14:52:52+0000

Answer:

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