Question

Point charges of 2x10^-9C are situated at each of the three corners of a square whose side is .20m. What would be the magnitude and direction of the resultant force on a point charge of -1x10^-9 if it were placed- at the vacant corner?

Answer 1

$F=F_1+F_2+F_3$

$F_1=k\frac{q_1q_2}{0.2^2}=9\cdot10^9\cdot\frac{2\cdot10^{-9}\cdot1\cdot10^{-9}}{0.2^2}=450\cdot10^{-9}(N)$

$F_2=F_1=450\cdot10^{-9}(N)$

$F_3=k\frac{q_1q_2}{(\sqrt{0.2^2+0.2^2})^2}=9\cdot10^9\cdot\frac{2\cdot10^{-9}\cdot1\cdot10^{-9}}{(\sqrt{0.2^2+0.2^2})^2}=$

$=225\cdot10^{-9}(N)$

$F_1+F_2=2F_1\cdot\cos45°=2\cdot450\cdot10^{-9}\cdot\cos45°=$

$=636\cdot10^{-9}(N)$

$F=636\cdot10^{-9}+225\cdot10^{-9}=861\cdot10^{-9}(N)$