Question

An infinite plane of charge with surface charge density $3.2\mu \mathrm{C}/{\mathrm{m}}^2$has a $20-cm$-diameter circular hole cut out of it. What is the electric field strength directly over the center of the hole at a distance of $12cm$?

Hint: Can you create this charge distribution as a superposition of charge distributions for which you know the electric field?

Short answer

Electrical field intensity directly over the middle of the outlet, $E_{\text{total}}=1.4\times {10}^5\frac{\mathrm{N}}{\mathrm{C}}$

Step-by-step solution

Electric field

When energy is present in any form, an electrical field is linked with each point in space.

The worth of E, also known as field of force strength, force field intensity, or just the electrical field, expresses the magnitude and direction of the electrical field.

Given values

Surface charge density$\eta =3.2\frac{\mu \mathrm{C}}{{\mathrm{m}}^2}$

${\epsilon }_o=8.85\times {10}^{-12}\frac{{\mathrm{C}}^2}{\mathrm{N}\times {\mathrm{m}}^2}$

$r=10\mathrm{cm}\to 0.10\mathrm{m}$

diameter is$z=12\mathrm{cm}\to 0.12\mathrm{m}$

Find field strength

Electric field strength is,

$E_{\text{total}}=E_p+E_d$

$E_{\text{total}}=\frac{\eta }{2{\epsilon }_o}-\frac{\eta }{2{\epsilon }_o}\left[1-\frac{z}{\sqrt{z^2+r^2}}\right]$

$E_{\text{total}}=\frac{\eta }{2{\epsilon }_o}\left[\frac{z}{\sqrt{z^2+r^2}}\right]$

$E_{\text{total}}=\frac{\left(3.2\times {10}^{-6}\frac{\mathrm{C}}{{\mathrm{m}}^2}\right)}{2\left(8.85\times {10}^{-12}\frac{{\mathrm{C}}^2}{\mathrm{N}·{\mathrm{m}}^2}\right)}\left[\frac{0.12\mathrm{m}}{\sqrt{(0.12\mathrm{m}{)}^2+(0.10\mathrm{m}{)}^2}}\right]$

$E_{\text{total}}=1.4\times {10}^5\frac{\mathrm{N}}{\mathrm{C}}$