91Ó°ÊÓ

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=$\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{∘}}=9.0×{10}^9 {\text{N.m}}^2/{\text{C}}^2$, Q=charge on the ring and a= radius of the ring

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

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

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

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).