Answer to Question #207193 in Electric Circuits for Hafee

Question

Calculate the electric field at one corner off a square 80𝑐𝑚 on a side if the other three corners are occupied by charges each of magnitude 18.2 x 10-4C.

Answer 1

Side of square, a = $80\space cm=80\times10^{-2}\space m$

Electric field due to horizontally placed charge, $E_1=\dfrac{kq}{a^2}=\dfrac{(9\times10^9)(18.2\times10^{-4})}{(80\times10^{-2})^2}$

$E_1=2.0475\times10^{9}\space N/C(left)$

Electric field due to vertically placed charge, $E_2=2.0475\times10^{9}\space N/C(downward)$

$E_{res_1}=\sqrt{E_1^2+E_2^2+2E_1E_2\cos90\degree}=3.50\times10^9\space N/C(diagonally)$

Electric field due to diagonally placed charge, $E_3=\dfrac{(9\times10^9)(18.2\times10^{-4})}{(80\sqrt{2}\times10^{-2})^2}$

$E_3=1.447\times10^{9}\space N/C$

$E_{res}=E_{res_1}+E_3=4.94\times10^{9}\space N/C(diagonally)$