Question

an electric dipole is kept on the axis of a uniformly charged ring at distance R/√2 from the centre of ring. the direction of dipole moment is along the axis. the dipole moment is P,charge of the ring is Q and radius of ring is R.calculate the force on the dipole??

Answer 1

electric field due to ring along its axis is given by, $E(x)=\frac{kQx}{(x^2+R^2)^{3/2}}$

where, x is the separation between centre of ring to observation point, Q is charge on ring and R is radius of the ring.

force on the dipole is given by, $F=P\frac{dE}{dx}$

so, first of all, differentiating electric field with respect to x,

e.g., $\frac{dE}{dx}=kQ\left[\frac{(x^2+R^2)^{3/2}-\frac{3}{2}(x^2+R^2)^{1/2}2x.x}{(x^2+R^2)^3}\right]$

=$kQ\sqrt{x^2+R^2}\left[\frac{x^2+R^2-3x^2}{(x^2+R^2)^3}\right]$

= $kQ\sqrt{x^2+R^2}\left[\frac{R^2-2x^2}{(x^2+R^2)^3}\right]$

at x = R/√2 , find value of dE/dx ;

dE/dx = $kQ\sqrt{x^2+R^2}\left[\frac{R^2-R^2}{(x^2+R^2)^3}\right]$

= 0

so, force on the dipole = P × 0 = 0

hence, when dipole is placed along axis of ring at R/√2 , then force acting on the dipole equals zero.