Question

q1 = 1nC, q2 = 1nC and q3 = -2nC. Which are located at x = 0 and y= 0, x= 1 cm and y= 0 and x= 1cm and y = 2cm respectively. What is the net force on q1 and in what direction?

Answer 1

Answer

force on q1 due to q_2

$F_1=\frac{Kq_1q_2}{r^2}$

$F_1=\frac{(9\times10^9)(1\times10^{-9})^2}{(0.01)^2}=0.00009N$

Force on q1 due to q3

$F_2=\frac{(9\times10^9)(1\times10^{-9})(2\times10^{-9})}{(\sqrt{5})^2}=8\times10^{-9}N$

AnAnd angle between F1 and F2 is 116.6° so net force

$F_{net}=\sqrt{0.00009^2+(8\times10^{-9})^2+2(0.00009)(8\times 10^{-9})cos116.6°}$

$F_{net}=9\times10^{-5}N$