Electric Field Intensity at a Point on the Axial line of An Electric Dipole (End on Position) - Param Himalaya

Questions: Derive an Expression for Electric Field Intensity at a Point on the Axial line of An Electric Dipole (End on Position) when the point is on the dipole axis

Solution: 

Consider an electric dipole consisting of charges +q and -q. The length of the dipole is 2l.

Consider a point P at a distance $r$ from the centre O of the dipole on the axial line of the dipole.

{STEP 1.} Electric field intensity at P due to $+q$ charge is given by : 

$\vec{E}_{+q} = \frac{1}{4\pi\epsilon_0} \frac{q}{BP^2} \quad \text{along (+) x-axis}$

$ = \frac{1}{4\pi\epsilon_0} \frac{q\hat{i}}{(r-l)^2}$

Electric field intensity at P due to $-q$ charge is given by

$\vec{E}_{-q} = \frac{1}{4\pi\epsilon_0} \frac{q}{AP^2} \quad \text{along (-) x-axis} $

$= \frac{1}{4\pi\epsilon_0} \frac{q(-\hat{i})}{(r+l)^2}$

$= \frac{-1}{4\pi\epsilon_0} \frac{q\hat{i}}{(r+l)^2}$

{STEP 2.} Net electric field intensity at point P due to electric dipole is given by

$\vec{E} = (\vec{E}_{+q}) + (\vec{E}_{-q})$

$= \left[ \frac{1}{4\pi\epsilon_0} \frac{q}{(r-l)^2} - \frac{1}{4\pi\epsilon_0} \frac{q}{(r+l)^2} \right] \hat{i}$

$ = \frac{q\hat{i}}{4\pi\epsilon_0} \left[ \frac{1}{(r-l)^2} - \frac{1}{(r+l)^2} \right]$

$\because (r-l).(r+l) = (r^2-l^2)$

$= \frac{q\hat{i}}{4\pi\epsilon_0} \left[ \frac{(r+l)^2 - (r-l)^2}{(r^2-l^2)^2} \right]$

$= \frac{q\hat{i}}{4\pi\epsilon_0} \left[ \frac{r^2 + l^2 + 2rl - r^2 - l^2 + 2rl}{(r^2 - l^2)^2} \right]$

$= \frac{q\hat{i}}{4\pi\epsilon_0} \left[ \frac{4rl}{(r^2 - l^2)^2} \right]$

$\vec{E}= \frac{1}{4\pi\epsilon_0} \frac{2r(q \times 2l)\hat{i}}{(r^2 - l^2)^2}$

{STEP 3.} But $(q \times 2l) = p$, where $p$ is the dipole moment

$\vec{E} = \frac{2rp\hat{i}}{4\pi\epsilon_0 (r^2 - l^2)^2} = \frac{2r\vec{p}}{4\pi\epsilon_0 (r^2 - l^2)^2} \quad (\because \vec{p} = p\hat{i}) \qquad \cdots (i)$

Here, direction of $p$ is along (+) x-axis, therefore, the direction of electric field intensity at a point on the axial line of the given electric dipole is along (+) x-axis. In other words, the direction of electric field at a point on the axial line of the electric dipole is along the direction of the electric dipole moment of the electric dipole. The magnitude of electric field intensity is given by

$E = \frac{2rp}{4\pi\epsilon_0 (r^2 - l^2)^2} \qquad \cdots (ii) $

Special case. If $r >> l$. That is, the distance of observation point P is at very-very large distance as compared to the dipole length or the dipole is very short, then $a^2$ can be neglected as compared to $r^2$. Hence, eqn. (ii) becomes

$E = \frac{2rp}{4\pi\epsilon_0 r^4}$

$ E = \frac{2p}{4\pi\epsilon_0 r^3} \qquad \cdots (iii)$

Thus, electric field due to an electric dipole at a point on its axial line is (i) directly proportional to the dipole moment ($p$) of the dipole and (ii) inversely proportional to cube of the distance of the point from the centre of the dipole.

Important Fact: 

Angle between the electric dipole moment $(\vec{p})$ and the electric field intensity $(\vec{E})$ due to the electric dipole at a point on its axial line is zero. Electric field is along the direction of dipole moment.