Identical point charges of +2.1 µc are fixed to three of the four corners of a square. what is the magnitude of the negative point charge that must be fixed to the fourth corner, so that the charge at the diagonally opposite corner experiences a net force of zero?

Suppose let’s say that the fourth corner with the negative charge D and the negative charge by -Q. Then the corner opposite D is B and the other corners are A and C. Corners A, B, and C each contains a positive charge q = 2.1 μC

Then we also call the side of square = S. Then by Pythagorean Theorem, the distance between B and D is:

S *sqrt(2)

So the force on the charge at B from the charge at A is:

f(A to B) = k q^2/s^2               

Then the force of charge at B from the charge at C is also:

f(C to B) = k q^2/s^2                 

Now if a charge -Q is to offer a force that counteracts both f ( C to B) and f(A to B), then the x component of that force is – f (A to B) and the y component is –f ( C to B). The force due to the charge - Q as force( D to B) so that:

force ( D to B) * cos (a) + force ( A to B) = 0

force ( D to B) * sin (a) + force ( C to B) = 0

where angle a = 45 degrees and hence sin (a) = cos(a) = sqrt(2)/2. Thμs we can solve for force ( D to B):

force (D to B) *cos(a) = -force (A to B)

force (D to B) = -force (A to B)/cos(a)

force (D to B) = -2*force (A to B)/sqrt(2)

force (D to B) = sqrt(2)*force ( A to B)

Now force ( D to B) = k*(-Q)*q/(2 S^2) so sμbstitμting:

k*(-Q)*q/(2S^2) = -sqrt(2)*k*q^2/S^2 --> -Q = 2*sqrt(2)q ---> Q = 2*sqrt(2)*q = 5.94 μC

5.94 μC

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary 

0.19 m/s and 0.45 m/s