Question

A short dipole is placed on the axis of a uniformly charged ring (total c distance $\frac{R}{\sqrt{2}}$ from centre of ring as shown in figure. Find the Force

Answer 1

0

Answer 2

The electric field on the axis of a uniformly charged ring is $E(x) = \frac{KQx}{(R^2 + x^2)^{3/2}}$. The force on a short dipole is $F = P \frac{dE}{dx}$. Differentiating $E(x)$ with respect to $x$ gives $\frac{dE}{dx} = KQ \left[ \frac{R^2 - 2x^2}{(R^2 + x^2)^{5/2}} \right]$. Substituting $x = \frac{R}{\sqrt{2}}$ into this derivative results in the numerator becoming $R^2 - 2\left(\frac{R^2}{2}\right) = R^2 - R^2 = 0$. Therefore, $\frac{dE}{dx} = 0$ at $x = \frac{R}{\sqrt{2}}$, which means the net force $F = P \cdot 0 = 0$.