Name: Electric Field Produced by a Dipole
Uploaded: 2019-07-19
Duration: 32 min 17 s
Description: What is the electric field of a dipole? Watch the video to understand how the vector addition can be used to determine the electric field of a dipole.

Question

Two fixed solid spheres of charges $Q$ and $k Q$ are separated by a distance $6 R .$ A charge $- q$ is to be projected along the line joining their centres, so that it can reach the surface of other solid sphere of charge $k Q .$

(i) If $k = 4,$ find minimum velocity to be given.
(ii) If we do not want to give any velocity, what should be the minimum value of $k ?$

Accepted Answer

First finding the point where Electric field is zero, after that point charge will automatically will move towards bigger sphere due to attraction force

Let $x$ be the point where field is zero

$\Rightarrow$ $\frac{k Q}{x^{2}} = \frac{4 k Q}{{(6 R - x)}^{2}}$ , here $k = \frac{1}{4 {πε}_{\circ}}$

$\Rightarrow 6 R = 3 x$

$\Rightarrow x = 2 R$

Using Energy conservation from point $1$ to that point where Electric field at point $(2)$ is zero

$\Rightarrow {(U_{Q} + U_{4 Q} + \frac{1}{2} m v^{2})}_{1} = {(U_{Q} + U_{4 Q})}_{2}$

$\Rightarrow \frac{- k Q q}{R} + \frac{- 4 k Q q}{5 R} + \frac{1}{2} m v^{2} = \frac{- k Q q}{2 R} + \frac{- 4 k Q q}{4 R}$

$\Rightarrow \frac{1}{2} m v^{2} = \frac{3 k Q q}{10 R}$

$\Rightarrow v = \sqrt{\frac{3 k Q q}{5 R}} = \sqrt{\frac{3 Q q}{20 {πε}_{\circ} mR}}$

If no velocity is given at point $1$ , then at point $1$ if filed is zero, then charge itself moves towards bigger sphere

Then for the field zero at point $1$

$\Rightarrow \frac{k Q}{R^{2}} = \frac{k K Q}{{(6 R - R)}^{2}}$

$\Rightarrow 25 R^{2} = K R^{2}$

$\Rightarrow K = 25$

Important Questions on Electrostatic Potential and Capacitance

Learn and Prepare for any exam you want !