Question

Calculate the magnitude of the electric field at one corner of a square 1.22 m on a side if the other three corners are occupied by 5.25×10−6 C charges.

Answer 1

$\bold E=\bold E_1+\bold E_2+\bold E_3$

$E_1=k\frac{q}{1.22^2}=9\cdot10^9\cdot \frac{5.25\cdot10^{-6}}{1.22^2}=31745(V/m)$

$E_3=k\frac{q}{1.22^2}=9\cdot10^9\cdot \frac{5.25\cdot10^{-6}}{1.22^2}=31745(V/m)$

$E_2=k\frac{q}{(1.22\cdot\sqrt2)^2}=9\cdot10^9\cdot \frac{5.25\cdot10^{-6}}{(1.22\cdot\sqrt2)^2}=15963(V/m)$

So,

$E=31745\cdot \cos45°+31745\cdot \cos45°+15963=60857(V/m)$