Question

Three charged particles are placed so that each particle is at the corner of a triangle that lies in the x-y plane. The charge at the origin is –3.0 μC, the charge at x = 6.0 cm on the x-axis is+4.0 μC, and the charge at y = 4.0 cm on the y-axis is +2.5 μC. What is the net electric force on the –3.0 μC charge?

Answer 1

Answer

Force on charge placed at origine

in x direction

$F_x=\frac{(9\times 10^9)(3\mu C) (4\mu C) }{(6cm)^2}=180N$

In Y direction

$F_x=\frac{(9\times 10^9)(3\mu C) (2.5\mu C) }{(4cm)^2}=168.7N$

Net force

$F_{net}=\sqrt{(180)^2+(168.7)^2}\\F_{net}=246.7N$