In two opposite vertices of the square, there are two identical charges of 800 nC, interacting with a force of 0.4 N

In two opposite vertices of the square, there are two identical charges of 800 nC, interacting with a force of 0.4 N. Determine the field strength in the center of the square, if another charge of 0.08 nC is placed in one of the free vertices of the square.

Let us find the value of the diagonal of the square, for this we use the Coulomb’s law:

F = 1 / 4πέέ0 * q1 * q2 / r ^ 2, έ is the dielectric constant of the medium, έ0 is the electrical constant, q1 and q2 are the magnitudes of the charges, r is the distance, then:

r ^ 2 = 1 / 4πέέ0 * q ^ 2 / F.

r = q √1 / 1 / 4πέέ0F;

r = 800 * 10 ^ (- 9) * √1 / 4 * 8.85 * 10 ^ (- 12) * 0.4 = 8 * 10 ^ (- 7) * √7 * 10 ^ (10) = 21 , 16 * 10 ^ (- 2) = 0.2 m.

The distance from the third charge to the center will be:

l = r / 2 = 0.1m

The tension is equal to:

E = 1 / 4πέέ0 * q / l.

E = 0.009 * 0.08 * 10 ^ (- 9) / 0.1 = 8 * 10 ^ (- 9)