Question

Three charges each equal to $+4nC$ are placed at the three corners of a square of side $2cm$. Find the electric field at the fourth corner.

Answer 1

$AD=2cm=2\times {10}^{-2}m$

$CD=2cm=2\times {10}^{-2}m$
$BD=2\sqrt{2}cm=2\sqrt{2}\times {10}^{-2}m$
$E_A=\frac{1}{4\pi {\epsilon }_0}\frac{Q^{}}{r^2}=9\times {10}^9\times \frac{4\times {10}^{-9}}{{(2\times {10}^{-2})}^2}=9\times {10}^{4}N{C}^{-1}$ along $AD$
$E_C=\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9\times \frac{4\times {10}^{-9}}{{(2\times {10}^{-2})}^2}=9\times {10}^{4}N{C}^{-1}$ along $CD$
$E_B=\frac{1}{4\pi {\epsilon }_0}\frac{Q_B}{r^2}=9\times {10}^9\times \frac{4\times {10}^{-9}}{{(2\sqrt{2}\times {10}^{-2})}^2}=4.5\times {10}^{4}N{C}^{-1}$ along $BD$

$X-$component of the field $E_x=E_A+E_B\cos 45=9\times {10}^4+4.5\times {10}^4\times 0.707$

$E_x=12.1815\times {10}^{4}N{C}^{-1}$

$Y-$component of the field $E_y=E_C+E_B\cos 45=9\times {10}^4+4.5\times {10}^4\times 0.707$
$E_x=12.1815\times {10}^{4}N{C}^{-1}$

Magnitude $E=\sqrt{E_x^2+E_y^2}=\sqrt{{(12.1815\times {10}^4)}^2+{(12.1815\times {10}^4)}^2}$
$E=\sqrt{2\times {(12.1815\times {10}^4)}^2}=17.2272\times {10}^{4}N{C}^{-1}$
OR
$E_{AC}=\sqrt{E_A^2+E_C^2}=\sqrt{{(9\times {10}^4)}^2+{(9\times {10}^4)}^2}$
$E_{AC}=\sqrt{2{(9\times {10}^4)}^2}=12.726\times {10}^{4}N{C}^{-1}$ along $BD$
$E=E_B+E_{AC}=4.5\times {10}^4+12.724\times {10}^4=17.226\times {10}^{4}N{C}^{-1}$ along $BD$