A circle with a radius of R is inscribed in a square. On the corner of the square are point charges of +q, +q, -q, and +q. On the circle are charges of +2q, +2q, -q, and +q. (See sketch.) If q = 4.8 micro-coulombs, and the total potential at the center of the circle due to all the charges on both the circle and the square is 2.5 x 103 V, what is the radius of the circle in meters?

To find the radius of the circle in meters, we need to use the given information to calculate the net potential at the center of the circle due to all the charges. Then we can relate that net potential to the electric field at the center of the circle and find the radius using the electric field equation for a point charge.

First, let's calculate the net potential at the center of the circle. The total potential at the center of the circle is given as 2.5 x 10^3 V (volts). This potential is the sum of the potentials due to the charges on both the circle and the square.

The potential due to a point charge is given by the equation:
V = k * (Q / r)

Where V is the potential in volts, k is Coulomb's constant (approximated as 9 x 10^9 Nm^2/C^2), Q is the charge in coulombs, and r is the distance from the charge to the point where we are calculating the potential.

Let's calculate the net potential due to the charges on the square first. The two positive charges (+q) on the corner of the square do not contribute to the net potential at the center of the circle since the distance between them and the center of the circle is zero. The negative charge (-q) on the corner of the square contributes a potential of -k * (q / r) due to its position.

Next, let's calculate the net potential due to the charges on the circle. The two positive charges (+2q) on the circle contribute a potential of 2 * k * (q / r) each, since the charges on the circle are twice the magnitude of the charges on the square. The negative charge (-q) on the circle contributes a potential of -k * (q / r), and the positive charge (+q) on the circle contributes a potential of k * (q / r).

Adding up the potentials from the square and the circle, we get:
- k * (q / r) + 2 * k * (q / r) + 2 * k * (q / r) - k * (q / r) + k * (q / r) = 2.5 x 10^3 V

Simplifying, we have:
3 * k * (q / r) = 2.5 x 10^3 V

Now, let's substitute the given values into the equation.
q = 4.8 micro-coulombs = 4.8 x 10^(-6) C (coulombs)
k = 9 x 10^9 Nm^2/C^2

Substituting these values, we have:
3 * (9 x 10^9 Nm^2/C^2) * (4.8 x 10^(-6) C / r) = 2.5 x 10^3 V

Simplifying further, we get:
(27 x 10^9 Nm^2/C^2 * 4.8 x 10^(-6) C) / r = 2.5 x 10^3 V

Combining the terms, we have:
(129.6 x 10^3 Nm^2/C) / r = 2.5 x 10^3 V

Now, let's solve for the radius (r) by isolating it on one side of the equation. We can do this by cross-multiplying and dividing:

(129.6 x 10^3 Nm^2/C) = (2.5 x 10^3 V) * r

r = (129.6 x 10^3 Nm^2/C) / (2.5 x 10^3 V)

r = 51.84 m^2/C * V

Therefore, the radius of the circle is approximately 51.84 meters.