Answer to Question #215469 in Electric Circuits for Tianye Kwadwo Fred

Question

Four electrical point charges with magnitude +100 form a square with a side of 2 m. find the magnitude and the direction of the net electric force acting at the bottom right of the square.

Answer 1

$F_{AB}=\frac{kq_1q_2}{r_{AB}^2}=\frac{k\times10^2\times10^2}{2^2}=\frac{k\times10^4}{4}$

$F_{CB}=\frac{kq_1q_2}{r_{CB}^2}=\frac{k\times10^2\times10^2}{2^2}=\frac{k\times10^4}{4}$

$F_{net}=F'_{net}+F_{DB}$

$F_{DB}=\frac{kq_1q_2}{r_{DB}^2}=\frac{k\times10^2\times10^2}{(2\sqrt2)^2}=\frac{k\times10^4}{8}$

$F'_{net}=\sqrt{F_{AB}^2+F_{CB}^2}=\sqrt{2}F_{AB}$

$F'_{net}=\sqrt{2}\frac{k\times10^4}{4}$

$F_{net}=F'_{net}+F_{DB}=(\sqrt{2}+\frac{1}{2})\frac{k\times10^4}{4}$

$F_{net}=(2\sqrt{2}+1)\frac{9\times10^{13}}{8}N$