Question

There are four charges, each with a magnitude of 2.16

Answer 1

Given:-

Area of square (A) = 0.33 m²

Side of square (S) = 0.165 m

Diagonal distance (d) = sqrt( S² + S²) =S*sqrt2 = 0.233 m

Charge is -ve and +ve So, we put the charge as given in the figure so that the net force on any charge is directed toward the center of the square.

Now each charge is attracted to the side charges, and repelled by the opposite charge across the diagonal.

Consider the charge at a corner being attracted to the two adjacent charges.  

The directions of these two forces are perpendicular to each other, so the resultant force is along the diagonal.

=> F1 = 2*sqrt2*k*q*q/S^2 (Repulsive force)

Consider the force across the diagonal due to the charge across the diagonal force

=> F2= k*q*q/d^2 (Repulsive force)

Net force toward the center:-

F= F1-F2

F= 2*sqrt2*k*q*q/S^2 - k*q*q/d^2

F= 9*10⁹* 2.16*10^-6*2.16*10^-6  [ 2*1.414/0.165² - 1/0.233² ]

F=41.9904 * 10^-3 [103.89 - 18.41]

F=41.9904 * 10^-3 * 85.47

F=3.5889 N

The magnitude of the net electrostatic force experienced by any charge is 3.5889 N.