Question

An infinite plane of charge has a surface charge density of 5 µC/m2 . How far apart are the equipotential surfaces whose potentials differ by 105 V? The permittivity of free space is 8.85 × 10−12 C 2 /N · m2 . Answer in units of mm.

Answer 1

Answer:

Distance in mm will be 0.3718 mm

Explanation:

We have given charge surface charge density $\rho _s=5\mu c/m^2=5\times 10^{-6}\mu c/m^2$

We know that electric field due to surface charge density is given by

$E=\frac{\rho _S}{2\epsilon _0}=\frac{5\times 10^{-6}}{2\times 8.85\times 10^{-12}}=2.824\times 10^5Volt/m$

We have given potential difference V = 105 volt

We know that potential difference is given by $V=Ed$

So $105=2.824\times 10^5\times d$

$d=37.181\times 10^{-5}m=0.3718mm$