Question:A ring-shaped conductor with radius a = 2.50 cm has a total positive charge Q = +0.125 nC uniformly distributed around it (see Fig. 21.23). The center of the ring is at the origin of coordinates O. (a) What is the electric field (magnitude and direction) at point P, which is on the x-axis at x = 40.0 cm? (b) A point charge q = -2.50 μC is placed at P. What are the magnitude and direction of the force exerted by the charge q on the ring?

Electric field at a distance x on the axis of a uniformly charged ring is

$\mathbf{E}\mathbf{=}\frac{\mathbf{k}Q\mathbf{x}}{{\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}\mathbf{)}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}$

Where k=$\frac{1}{4\pi {\epsilon }_{\circ }}=9.0\times {10}^9 {\text{N.m}}^2/{\text{C}}^2$, Q=charge on the ring and a= radius of the ring

$$\begin{gathered} E=\frac{9\times {10}^9\times 0.125\times {{10}^{-}}^9\times 0.4}{{({0.025}^2+{0.4}^2)}^{3/2}} \\ =\frac{0.45}{{0.160}^{3/2}} \\ =\text{7.0}N/C \end{gathered}$$

Now as because the ring is positively charged so direction of electric field at point P away from it. Which means the direction of electric field is towards positive x-axis

Now as per we know that force on charge q placed in an electric field E is

F = qE

$$\begin{gathered} F=2.50\times {10}^{-6}C\times 6.99N/C \\ =1.74\times {10}^{-5}N \end{gathered}$$

Hence the magnitude of force exerted by the charge q on the ring is1.74 x 10^-5 N.

Now according to Newton's 3rd law charge q will also exert a force on the ring which is equal in magnitude but opposite in direction.

The direction of force exerted by charge q on the ring will be towards positive X-axis and towards charge q, because the force on the ring due to change q will be opposite to the force exerted by the ring on the charge q (which is towards negative x-axis direction).