Question

A +200 uC charge is placed at the origin. A +400 uC charge is placed at 2m above the origin(y-axis), and a -600 uC charged is placed 3m to the right of the origin (x-axis). What is the magnitude and direction of the net force acting on the charge placed at the origin?

Answer 1

Answer

Force due to charge +400uC

$F_1=\frac{9\times 10^9\times200\times400\times 10^{-12}}{4}\\=180(-j) N$

Force due to charge -600uC

$F_1=\frac{9\times 10^9\times200\times600\times 10^{-12}}{9}\\=120(i) N$

So their resultant

$। F। =\sqrt{F^2_1+F^2_2}\\=\sqrt{(180)^2+(120)^2}$

$=216.33N$

Direction

F=216.33$(i-j) /\sqrt{2}$ N