Answer to Question #196455 in Electric Circuits for Saliu Ibrahim

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

Gives

r=80cm 0.80m

Charge=18.2$\times20^{-4}C$

$E_1=\frac{Kq_1q_2}{r^2}$

$E_2=\frac{Kq_1q_2}{r^2}$

$E_3=\frac{Kq_1q_2}{r'^2}$

E₁And E₂ perpendiculer

Resultant

E'=$\sqrt{E_1^2+E_2^2}$

$E'=\sqrt2E_1=\sqrt{2}E_2$

$E_3=\frac{kq_1q_2}{2r^2}$

Where r'=$\sqrt{2}r$

$E_1=\frac{9\times10^9\times18.2\times10^{-4}\times18.2\times10^{-4}}{0.8^2}$

$E_1=46.59\times10^{3}N/C$

$E_1=E_2=46.59\times10^3N/C$

$$E'=\sqrt2\times46.59\times10^3N/C=65.88\times10^3N/C$$

1. 
   

$E_3=\frac{9\times10^9\times18.2\times10^{-4}\times18.2\times10^{-4}}{(\sqrt2\times0.8)^2}$

$E_3=23.295\times10^3N/C$

$E_{net}=E'+E_3$

$$E_3=65.88\times10^3+23.295\times10^3=89.175\times10^3 N/C$$