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

Answer:

so the distance between two points are

$d = 0.246 \times 10^{-3} m$

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

$E = \frac{\sigma}{2\epsilon_0}$

now we have

$E = \frac{7.2 \times 10^{-6}}{2(8.85 \times 10^{-12})}$

$E = 4.07 \times 10^5 N/C$

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

$E.d = \Delta V$

$4.07 \times 10^5 (d) = 100$

$d = \frac{100}{4.07 \times 10^5}$

$d = 0.246 \times 10^{-3} m$