Question

Q5.A dipole having dipole moment 'p' is place at the axis of a charged ring having charge Q and radius R at a distance x from the center as shown in figure.Find the force experienced by the dipole of dipole moment p^(rarr)=phat imath placed along the axes of the uniformly charged ring.​

Answer 1

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Answer 2

The force experienced by the dipole of dipole moment is F = $\frac{KpQ}{3\sqrt{3}R }$

Given:

Charge of the ring = Q

The radius of the ring = R

The dipole moment of the dipole is given p

The separation between the ring and the dipole is x

To find:

The force experienced by the dipole of dipole moment.

Solution:

The electric field due to the ring along the axis can be represented by

E(x) = $\frac{KQx}{(x^{2}+R^{2})^{3/2} }$

Force on dipole is given by F = dE/dt.

Now, differentiating the electric field with respect to x

dE/dt = KQ $\sqrt{x^{2}+R^{2} }$  $\frac{R^{2}-2x^{2} }{(x^{2} +R^{2})^{3} }$

Now, at x = $\sqrt{2}$R dE/dt is

dE/dt = KQ $\sqrt{2R^{2}+R^{2} }$  $\frac{R^{2}-2(\sqrt{2} )^{2} }{[\sqrt{2R)} ^{2} +R^{2}]^{3} }$

dE/dt = KQ $\sqrt{3}$R²  $\frac{-3R^{2} }{27R^{5} }$

dQ/dt = -KQ $\frac{\sqrt{3} }{9R}$

dQ/dt = -KQ $\frac{1}{3\sqrt{3}R }$

So, the force on the dipole is:

F = p dE/dt

F = $\frac{KpQ}{3\sqrt{3}R }$

The force experienced by the dipole of dipole moment is F = $\frac{KpQ}{3\sqrt{3}R }$

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