# Electric Field At An Axial Point Of A Dipole | cbse24.com

### The electric field at an axial point of an electric dipole:

As shown in figure an electric dipole consisting of charge +q and -q , separated by distance 2a and placed in a vacuum . Let P be a point on the axial line at distance r from the center O of the dipole on the side of the charge +q.

 The electric field at an axial point of a dipole

By this picture, you can see the resultant electric field at an axial point of a dipole this value is 3.51 V/m but if you start increasing distance mean point p on the axial point of dipole than field start decreasing.

 The electric field at an axial point of a dipole at a distance

so here you can see that if I increase the distance P on the axial point of a dipole than the electric field start decreasing and its value will be  1.72 V/m. so what I found? , with increasing the distance the electric field will start decreasing.

 The electric field at an axial point of a dipole at a far distance

### Calculation of the net electric field at an axial point of a dipole:

1. Electric field due to charge  -q at point P is

$\overrightarrow{E{-q}}=\frac{-q}{4\pi \varepsilon _{0}(r+a)^{2}}\hat{p}$       ..................(towards left)

2. Electric field due to charge  +q at point P is

$\overrightarrow{E{+q}}=\frac{+q}{4\pi \varepsilon _{0}(r-a)^{2}}\hat{p}$   .........................(towards the right)

Note- Where  $\hat{p}$  is the unit vector along the dipole axis from -q to +q

Hence the resultant electric field at point P is

$\overrightarrow{E_{axial}}=\overrightarrow{E_{-q}}+\overrightarrow{E_{+q}}$

$\overrightarrow{E_{axial}}=\frac{q}{4\pi \varepsilon _{0}}\times\left ( \frac{1}{(r-a))^{2}}-\frac{1}{(r+a))^{2}} \right )\hat{p}$

$=\frac{q}{4\pi \varepsilon _{0}} .\frac{4ar}{(r^{2}-a^{2})^{2}}\hat{p}$

or          $\overrightarrow{E_{axial}}=\frac{1}{4\pi \varepsilon _{0}} .\frac{2pr}{(r^{2}-a^{2})^{2}}\hat{p}$

Here   P= q x 2a = dip[ole moment .
for  r>>a  , a2  can be neglected compared to r2

$\overrightarrow{E_{axial}}=\frac{1}{4\pi \varepsilon _{0}} .\frac{2p}{r^{3}}\hat{p}$   .......................(towards right according to vector addition)

Note- The electric field at any point of the dipole acts along the dipole axis from negative to the positive charge i.e in the direction of dipole moment $\overrightarrow{p}$.

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