Question

A charge of 9.00 mC is placed at each corner of a square 0.310 m on a side.

Determine the magnitude of the force on each charge.

Answer 1

First find the force exerted by each of the charges on the adjacent corners and their sum.
According to Coulomb's law the magnitude of the force is
F1 = F2 = kd · Q² / r²
= 8.988×10+9 Nm²/C² · (9.0×10-3C)² / (0.31m)²
= 7575.733611kN
These two forces acting perpendicular to each other.
So they can easily added. The magnitude of the sum vector of two perpendicular vectors is the the square root of the sum of their squared magnitudes. (you can easily show this by a sketch)
Hence the magnitude of the sum force is:
F12 = √( F1² + F2²)
= √2 · 159.3kN
= 10713kN
Because F1 and F2 are equal F12 acts along the diagonal away from the center of the square.

The third force, exerted by the charge on the opposite corner acts in that direction, too. The distance between the charges along diagonal can be found from Pythagorean theorem:
r² = (0.31m)² + (0.31m)²
Hence the magnitude of the force is:
F3= 8.988×10+9 Nm²/C² · (9.0×10-3C)² / [2· (0.31m)² ]
= 3787.86kN

Now have two forces acting in the same direction. Magnitude of their sum vector is equal to the sum of their magnitudes. Thus the magnitude of the force acting on a charge is:
F123 = F12 + F3
=10713kN +3787.8kN
= 14500.86kN