Question

Find the magnitude of the net electrostatic force exerted on the point charge q2 in the figure. Let q = +2.1 µC and d = 39 cm.

Answer in Newtons

Answer 1

We will first find between q2 and q3,
F=kq2q3/d^2
= (9 x 10^9)(2 x 10^-6)(3 x 10^-6) / (.39)^2
= .355 N to the left, repulsive force
now between q2 and q1
F=(9 x 10^9)(2 x 10^-6)(1 x 10^-6) / (.39)^2
= .118 N down, attractive force
between q2 and q4, with the distance d√2 due to pythagorean theorem

F=(9 x 10^9)(2 x 10^-6)(4 x 10^-6) / (.39 x root 2)^2

= .237 N diagonal left up, repulsive force

We have to break down the last one into x and y components, which will be the same because the angle is 45

sin45 = x/.237

x= .167 N up

cos 45 = x/.237

x= .167 N left

Now add y components, .167 + (-.118) = .050 N up

x components, .335 + .167 = .502 N to left

use pythagorean theorem to get magnitude

(.050)^2 + (.502)^2 = x^2

x= .504 N is magnitude, and if u need direction, then use trig