Electric Field due to a Dipole

There are three positions for which we will derive the expression for electric field due to a dipole:

1.At axial point

2. At equatorial point

3. At any point

Electric Field due to a Dipole at its Axial Point

Let us consider an electric dipole of dipole moment \(\vec{p}\) and length 2\(\vec{l}\). Let P be a point on the axis of the dipole, at a distance r from the centre of the dipole.

The distance of the point P from +q charge= BP = (r-l)

The distance of the point P from -q charge= AP = (r+l)

Electric Field at P due to +Q charge

\[E_1 = \frac{k \vert q \vert}{BP^2}\]

\[E_1= \frac{k \vert q \vert}{(r- l)^2}\]

Here k is Electrostatic Force Constant or Coulomb’s Constant.

The direction of \(E_1\) is along \(\vec{OP}\)

Electric Field at P due to -q charge

\[E_2 = \frac{k \vert q \vert}{AP^2}\]

\[E_2= \frac{k \vert q \vert}{(r+l)^2}\]

The direction of \(E_1\) is along \(\vec{PO}\)

According to the Superposition Principle the net electric field will be the vector sum of \(E_1\) and \(E_1\) , which in this case equals  \(E_1\) minus \(E_2\) since they have opposite directions.

Hence the net electric field is given by

$$
\begin{align*}
E &= E_1 – E_2 \\
  &= \frac{k|q|}{(r – l)^2} – \frac{k|q|}{(r + l)^2} \\
  &= k|q| \left( \frac{1}{(r – l)^2} – \frac{1}{(r + l)^2} \right) \\
  &= k|q| \left( \frac{(r + l)^2 – (r – l)^2}{(r – l)^2 (r + l)^2} \right) \\
  &= k|q| \left( \frac{4rl}{(r^2 – l^2)^2} \right) \\
  &\quad \boxed{Since (r + l)^2 – (r – l)^2 = 4rl} \\
  &= \frac{2k|q|(2l)r}{(r^2 – l^2)^2} \\
  &= \frac{2kpr}{(r^2 – l^2)^2} \quad \text{[since } p = |q|(2l) \text{]} \\
\Rightarrow\quad
\boxed{E = \frac{2kpr}{(r^2 – l^2)^2}}
\end{align*}
$$

For a short dipole

\(\ r^2>> l^2\)

Since \(\ r^2 + l^2 \approx r^2\), we have:
\[ E = \frac{2kpr}{(r^2)^2} \]

\[\Rightarrow \quad E = \frac{2kpr}{r^4}\]

\[\Rightarrow \quad \boxed {E = \frac{2kp}{r^3}}\]

It is evident from the diagram that the direction of electric field at any point on the axial line is same as the direction of the dipole moment. Therefore in vector terms we can write the expression as

\[\vec{E} = \frac {2k \vec{p}r}{ ( r^2- l^2)^2}\]

For a short dipole

\[\vec{E} = \frac {2k\vec{p}r}{r^3}\] 

Conclusion:

1. The direction of axial field is same as the dipole moment vector.
2. \(E(axial) = \Large \frac{2kp}{r^3}\)

Electric Field due to a Dipole at its Equatorial Point

Let us consider an electric dipole of dipole moment \(\vec{p}\) and length 2\(\vec{l}\).Let P be a point on the equatorial line at a distance r from the centre of the dipole.

As the point P lies on the perpendicular bisector of the line AB, the distance of the point P from both the charges will be equal.

i.e. AP = BP =\(\sqrt{r^2+l^2}\)

Now,

Electric Field at P due to +q charge

\[E_1= \frac{k \vert q \vert}{BP^2}\]

\[\Rightarrow E_1=\frac{k \vert q \vert}{r^2+l^2}\]

It’s direction is along \(\vec{BP}\)

Electric Field at P due to -q charge

\[E_2= \frac{k\vert q \vert}{AP^2}\]

\[\Rightarrow E_2=\frac{k \vert q \vert}{r^2+l^2}\]

It’s direction is along \(\vec{PA}\)

Since the magnitudes of \(\vec{E_1}\) and \(\vec{E_2}\) are equal. Therefore, when we resolve them into their respective components as shown in figure, their sine components cancel each other and cosine components add up to give the resultant.

Hence,

\[
\begin{aligned}
E &= E_1 \cos\theta + E_2 \cos\theta \\
  &= 2E_1 \cos\theta \\
  &= 2 \cdot \frac{k |q|}{(r^2 + l^2)} \cos\theta \\
  &= 2k|q| \cdot \frac{l}{\sqrt{r^2 + l^2}} \cdot \frac{1}{(r^2 + l^2)} \\
  &= \frac{2k|q|l}{(r^2 + l^2)^{3/2}} \\
  &= \frac{kp}{(r^2 + l^2)^{3/2}} \\
\Rightarrow \boxed{E = \frac{kp}{(r^2 + l^2)^{3/2}}}
\end{aligned}
\]

For a short dipole

\[
\begin{aligned}
&\text{Since } r^2 \gg l^2 \\
&\Rightarrow r^2 + l^2 \approx r^2 \\
\\
E &= \frac{kp}{(r^2 + l^2)^{3/2}} \\
&\approx \frac{kp}{r^3} \\
\\
\Rightarrow\boxed{E = \frac{kp}{r^3}}
\end{aligned}
\]

It is evident from the diagram that the direction of electric field is opposite to that of dipole moment.Therefore,

\[
\boxed{
\vec{E} = -\frac{k \vec{p}}{(r^2 + l^2)^{3/2}}
}
\]
For a short dipole
\[
\boxed{
\vec{E} = -\frac{k \vec{p}}{r^3}
}
\]

Conclusion :

1. The direction of equatorial field is opposite to that of dipole moment.
2. \(E(equatorial)= \Large\frac {kp}{r^3}\)

Electric Field Due to a Dipole at Any Point

Let us consider an electric dipole of dipole moment \(\vec{p}\) and length ‘2l’.Let P be a point at a distance r from the centre O of the dipole as shown in figure.

We choose x and y axes as shown in figure and resolve the dipole moment \(\vec{p}\) into x and y components. For the x-component \(p_x\) the point P lies on the axial line and for the y-component \(p_y\) the point P lies on the equatorial line. Therefore

Electric Field at P due to \(p_x\)

\[
E_x = \frac{2kp_x}{r^3}
\]

\[
\Rightarrow E_x = \frac{2kp \cos\theta}{r^3} \quad \left[ \because p_x = p \cos\theta \right]
\]

Electric Field at P due to \(p_y\)

\[
E_y = \frac{-k p_y}{r^3}
\]

\[
= \frac{-k(-p \sin\theta)}{r^3}
\]

\[
\Rightarrow E_y = \frac{k p \sin\theta}{r^3}
\]