A charge of +2q is fixed to one corner of a square, while a charge of -3q is fixed to the diagonally opposite corner. Expressed in terms of q, what charge should be fixed to the center of the square, so the potential is zero at each of the two empty corners?

Potential is a scalar, so you can just add all four contributions.

Vt= -3q/r + 2q/r + 2X/r=0

solve for charge X

To find the charge that should be fixed to the center of the square so that the potential is zero at each of the two empty corners, we need to calculate the potential at those corners.

First, let's label the corners of the square. We will call the corner with the +2q charge "A," the corner with the -3q charge "B," and the remaining two empty corners "C" and "D."

The potential at any point in space due to a point charge is given by the equation:

V = k * (q / r)

Where V is the potential, k is the electrostatic constant, q is the charge, and r is the distance between the point charge and the point where the potential is being calculated.

Since the potential at the empty corners C and D should be zero, the potential due to these corners must cancel out. Therefore, the potential at C due to the charge at A should be equal in magnitude and opposite in sign to the potential at D due to the charge at B.

Let's calculate these potentials:

The potential at corner C due to the charge at corner A is given by:

Vca = k * (2q / r)

And the potential at corner D due to the charge at corner B is given by:

Vdb = k * (-3q / r')

Since the magnitude of the potentials should be the same, we can equate these two equations:

k * (2q / r) = k * (-3q / r')

We can cancel out the electrostatic constant k and simplify the equation:

2q / r = -3q / r'

Cross-multiplying, we get:

2q * r' = -3q * r

Dividing both sides by q:

2 * r' = -3 * r

Now we can solve for r' in terms of r:

r' = (-3 * r) / (2)

To find the total distance r', we can use the Pythagorean theorem for the diagonal of the square, which is equal to the side length of the square multiplied by the square root of 2:

r' = s * sqrt(2)

Therefore, we have:

s * sqrt(2) = (-3 * r) / (2)

Simplifying further:

s = (-3 * r) / (2 * sqrt(2))

Now, knowing that the charge of the center of the square should make the potentials at C and D zero, we can calculate the potential at these corners due to the charge at the center.

The potential at corner C due to the charge at the center is given by:

Vcc = k * (qc / rc)

And the potential at corner D due to the charge at the center is given by:

Vcd = k * (qc / rd)

Since the potentials at C and D should be zero, we have:

Vcc = 0

And

Vcd = 0

Therefore, we can write:

k * (qc / rc) = 0

And

k * (qc / rd) = 0

Since both of these equations should be zero, the charges qc and qc must be zero.

Thus, the charge that should be fixed to the center of the square so that the potential is zero at each of the two empty corners is zero.