Question

Derive an expression for the electric field at a point on the equatorial plane of an electric dipole.

Answer 1

Dipole is a pair of equal and opposite charges separated by some distance. Dipole moment of a dipole is the product of the charge with the separation distance between two opposite charges. To derive an electric field due to a dipole, first find the field for individual charges then take the resultant.

Formulae used:
Electric field at a distance d created due to a charge q is given by:
$\vec E = \dfrac{1}{{4\pi \varepsilon }}\dfrac{q}{{{d^2}}}\hat d$............(1)
Where,
$\vec E$ denotes the electric field vector,
$\varepsilon$ is permittivity of the medium,
$\hat d$ is the unit vector along the distance.
Dipole moment of a dipole is given by:
$\vec p = q\vec L$...................(2)
Where,
$\vec p$ denotes dipole moment,
q is the value of each charge,
$\vec L$ denotes the distance vector between the charges.

Complete step by step solution:
Step 1
Now, assume that the point is at a distance r away from the midpoint of the dipole. Also, take the distance between two charges as $L = 2l$. Hence, you’ll get d as: $d = \sqrt {{r^2} + {l^2}}$. Now, use these values in eq.(1) to get the electric field at P for the positive charge (+q) as:
${\vec E_1} = \dfrac{1}{{4\pi \varepsilon }}\dfrac{q}{{{{\left( {\sqrt {{r^2} + {l^2}} } \right)}^2}}}\hat d = \dfrac{1}{{4\pi \varepsilon }}\dfrac{q}{{\left( {{r^2} + {l^2}} \right)}}\hat d$
Notice, the direction of the electric field is in the direction of the distance i.e. outward from the charge.

Step 2
Similarly, as you did above, now find the electric field at P due to the negative charge (-q) as:
${\vec E_2} = \dfrac{1}{{4\pi \varepsilon }}\dfrac{{ - q}}{{{{\left( {\sqrt {{r^2} + {l^2}} } \right)}^2}}}\hat d = - \dfrac{1}{{4\pi \varepsilon }}\dfrac{q}{{\left( {{r^2} + {l^2}} \right)}}\hat d$
Here, the negative sign implies that the direction of electric field is opposite to the direction of the distance i.e. inward to the charge.

Step 3
Now, both the electric fields are at an angle $\theta$ w.r.t horizontal line (see the figure). Hence, the resulting vertical component of ${\vec E_1}$ and ${\vec E_2}$ is given by:
${{\vec E}_V} = {{\vec E}_1}\sin \theta + {{\vec E}_2}\sin \theta \\\ = \left( {\dfrac{1}{{4\pi \varepsilon }}\dfrac{q}{{\left( {{r^2} + {l^2}} \right)}} - \dfrac{1}{{4\pi \varepsilon }}\dfrac{q}{{\left( {{r^2} + {l^2}} \right)}}} \right)\sin \theta \hat v \\\ = 0 \\\$
So, the resulting electric field will be in a horizontal direction.
Hence, the value of the resulting electric field will be:
$E = {E_H} = {E_1}\cos \theta + {E_2}\cos \theta$.............(3)

Step 4
From the diagram we get, $\cos \theta = \dfrac{l}{{\sqrt {{r^2} + {l^2}} }}$and using eq.(2) we get the value of dipole moment as: $p = q \times 2l = 2lq$. Using these values in eq.(3) you’ll get:
And its direction will be horizontally from right to left direction (shown in figure).

E = \left( {\frac{1}{{4\pi \varepsilon }}\frac{q}{{\left( {{r^2} + {l^2}} \right)}} + \frac{1}{{4\pi \varepsilon }}\frac{q}{{\left( {{r^2} + {l^2}} \right)}}} \right) \times \frac{l}{{\sqrt {{r^2} + {l^2}} }} \\\ \therefore E = \frac{1}{{4\pi \varepsilon }} \times \frac{{2q}}{{\left( {{r^2} + {l^2}} \right)}} \times \frac{l}{{\sqrt {{r^2} + {l^2}} }} = \frac{1}{{4\pi \varepsilon }}\frac{p}{{{{\left( {{r^2} + {l^2}} \right)}^{\tfrac{3}{2}}}}} \\\ \end{gathered} $$ Step 5 As the charges in real dipoles stays quite close to each other so the distance r is effectively very large compared to 2l ($r \gg 2l$) so in the denominator we can neglect the term ${l^2}$ in comparison to ${r^2}$. Hence, the value of electric field becomes: $E = \frac{1}{{4\pi \varepsilon }}\frac{p}{{{{\left( {{r^2}} \right)}^{\tfrac{3}{2}}}}} = \frac{1}{{4\pi \varepsilon }}\frac{p}{{{r^3}}}$ **Note:** While doing the calculation many students make mistakes in obtaining the direction of the electric field produced by each charge. Specifically, for the negative charge the direction of the electric field will be just opposite. Also, while taking the approximation in the end one must ensure that the observation point must be far from the dipole.