Question

Three charges, each of magnitude 2 nC, are at separate corners of a square of edge length 2 cm. The two charges at opposite corners are positive, and the other charge is negative. Find the force exerted by these charges on a fourth charge q +3 nC at the remaining (upper right) corner. (Assume the +x axis is directed to the right and the ty axis is directed upward.)

Answer 1

`F_(net) = 1.23 * 10^(-4) N`

Step-by-step explanation:

Force due to two positive charges on 3 nC is given as

`F = (kq_1q_2)/(r^2)`

`F_1 = F_2 = ((9* 10^9)(2nC)(3 nC))/(0.02^2)`

`F_1 = F_2 = 1.35 * 10^(-4) N`

now the force on 3 nC charge due to opposite charge - 2 nC is given as

`F_3 = ((9* 10^9)(2nC)(3 nC))/(2(0.02)^2)`

`F_3 = 0.675 * 10^(-4) N`

now the force F3 is an attractive force while F1 and F2 is repulsive nature force

so we will have

`F_(net) = \sqrt2 F_1 - F_3`

`F_(net) = \sqrt2 (1.35 * 10^(-4)) - (0.675 * 10^(-4))`

`F_(net) = 1.23 * 10^(-4) N`