Question

In Fig. 21.10, a conducting ring 0.71 m in radius carries a charge of +580 nC. A point charge Q is placed at the center of the ring. The electric field is equal to zero at field point P, which is on the axis of the ring, and 0.73 m from its center. The point charge Q, in nC, is closest to:

-300

	210
	-210
	-420
	300

Answer 1

The radius of the conducting ring is R = 0.71 m
The charge carried by the ring is q = +580 nC = +580 * 10^-9 C
The electric field on the axis of the ring is
E = [k|qz|] / [(z^2 + R^2)^3/2]
where k = (1/4pieo) = 9 * 10^9 Nm^2/C^2 and z = 0.73 m
The electric force experienced by the point charge Q is
F = E * Q
The electric force is equal to the electric force between the charged ring and the point charge Q,therefore,we get
F1 = k * (q * Q/R^2)
Here,F = F1
or F = k * (q * Q/R^2)
or Q = (F * R^2/k * q)