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

Derive the expression for electric field intensity at a point on equatorial plane due to electric dipole.

Accepted Answer

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:
${\vec{E}}_{1} = \frac{1}{4 π ε} \frac{q}{{(\sqrt{r^{2} + l^{2}})}^{2}} \hat{d} = \frac{1}{4 π ε} \frac{q}{(r^{2} + l^{2})} \hat{d}$
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:
${\vec{E}}_{2} = \frac{1}{4 π ε} \frac{- q}{{(\sqrt{r^{2} + l^{2}})}^{2}} \hat{d} = - \frac{1}{4 π ε} \frac{q}{(r^{2} + l^{2})} \hat{d}$
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 $θ$ w.r.t horizontal line (see the figure). Hence, the resulting vertical component of ${\vec{E}}_{1}$ and ${\vec{E}}_{2}$ is given by:

${\vec{E}}_{v} = {\vec{E}}_{1} \sin θ + {\vec{E}}_{2} \sin θ = (\frac{1}{4 π ε} \frac{q}{{(\sqrt{r^{2} + l^{2}})}^{2}} - \frac{1}{4 π ε} \frac{q}{(r^{2} + l^{2})}) \sin θ \hat{v} = 0$

So, the resulting electric field will be in a horizontal direction.
Hence, the value of the resulting electric field will be:
$E = E_{H} = E_{1} \cos θ + E_{2} \cos θ$ .............(3)

Step 4
From the diagram we get, $\cos θ = \frac{l}{\sqrt{r^{2} + l^{2}}}$ And using eq.(2) we get the value of dipole moment as: $p = q \times 2 l = 2 l q$ . Using these values in eq.(3) you’ll get:
And its direction will be horizontally from right to left direction (shown in figure).
$E = (\frac{1}{4 π ε} \frac{q}{{(\sqrt{r^{2} + l^{2}})}^{2}} - \frac{1}{4 π ε} \frac{q}{(r^{2} + l^{2})}) \times \frac{l}{(r^{2} + l^{2})} E = (\frac{1}{4 π ε} \times \frac{2 q}{(r^{2} + l^{2})}) \times \frac{l}{(r^{2} + l^{2})} = (\frac{1}{4 π ε} \times \frac{p}{{(r^{2} + l^{2})}^{\frac{3}{2}}})$

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 ( $r ≫ 2 l$ ) so in the denominator we can neglect the term $l^{2}$ In comparison to $r^{2}$ . Hence, the value of electric field becomes:
$E = \frac{1}{4 π ε} \frac{p}{{(r^{2})}^{\frac{3}{2}}} = \frac{1}{4 π ε} \frac{p}{r^{3}}$

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.

Derive the expression for electric field intensity at a point on equatorial plane due to electric dipole.

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