Question

A dipole having charges +q and -q is fixed on a rough surface as shown.Another charge Q of mass m is placed at distance d from this dipole.Find the minimum value of d for which charge Q will not move.(given a<<

Answer 1

Given:

A dipole having charges +q and -q is fixed on a rough surface as shown.Another charge Q of mass m is placed at distance d from this dipole.

To find:

Minimum value of d for which charge Q will not move.

Calculation:

For equilibrium of the charge Q , the electrostatic force exerted by the dipole on the charge will be E equal and opposite to the frictional force exerted by the rough surface.

Let k be Coulomb's Constant, P be Dipole Moment :

$\therefore \: \rm{Electrostatic \: Force = Friction}$

$\rm{ = > \bigg ( \dfrac{2kP}{ {d}^{3} } \bigg)Q = \mu(mg)}$

$\rm{ = > \bigg ( \dfrac{2kP}{ \mu mg} \bigg)Q = {d}^{3} }$

$\rm{ = > {d} = \: \sqrt[3]{\bigg ( \dfrac{2kP}{ \mu mg} \bigg)Q} }$

Putting value of Dipole Moment :

$\rm{ = > {d} = \: \sqrt[3]{\bigg ( \dfrac{2(2qa)}{ 4\pi\epsilon_{0} \mu mg} \bigg)Q} }$

$\rm{ = > {d} = \: \sqrt[3]{\bigg ( \dfrac{qa}{ \pi\epsilon_{0} \mu mg} \bigg)Q} }$

So , final answer :

$\boxed{ \red{ \bold{ {d} = \: \sqrt[3]{\bigg ( \dfrac{qa}{ \pi\epsilon_{0} \mu mg} \bigg)Q} }}}$