Question

With the help of a labelled diagram, obtain an expression for the electric field intensity E at any point P on the equatorial line (broad-side-on position) of an electric dipole.

Accepted Answer

Expression for electric field intensity at a point due to dipole in equatorial position or broad-side on position.

Let $A B$ is an electric dipole of dipole length $2 l$ and dipole moment $P$ as $P = 2 lq$ , From the middle point of dipole at distance $r$ on equatorial line $H$ be a point where electric field intensity is to be determined.

According to figure

$OA = OB = ℓ, OH = r$ , then In $∆ HOA, AH = \sqrt{{OA}^{2} + {OH}^{2}} = \sqrt{r^{2} + l^{2}}$

Similarly $BH = \sqrt{r^{2} + l^{2}}$

At point two electric fields ${\vec{E}}_{HB}$ and ${\vec{E}}_{HA}$ are inclined at angle $2 θ$

By using the relation $E = \frac{1}{4 π ϵ_{0}} \cdot \frac{q}{r^{2}}$ for electric field intensity at a point due to source charge $q$

So, Electric field intensity at point $H$ due to $+ q$ charge

$= E_{HB} = \frac{1}{4 π ε_{0}} \cdot \frac{q}{(BH)^{2}} = \frac{1}{4 π ε_{0}} \cdot \frac{q}{{(\sqrt{r^{2} + l^{2}})}^{2}}$ $= \frac{1}{4 π ϵ_{0}} \cdot \frac{q}{r^{2} + l^{2}}$

And electric field intensity at point $H$ due to $- q$ charge

$= E_{HA} = \frac{1}{4 π ϵ_{0}} \cdot \frac{| - q |}{(AH)^{2}} = \frac{1}{4 π ϵ_{0}} \cdot \frac{q}{{(\sqrt{r^{2} + l^{2}})}^{2}}$ $= \frac{1}{4 π ε_{0}} \cdot \frac{q}{r^{2} + l^{2}}$

According to figure angle between ${\vec{E}}_{HB}$ and ${\vec{E}}_{HA}$ be $2 θ$

$∴$ By using parallelogram law of vector addition in parallelogram $HRTS$

$E_{net} = \sqrt{{E^{2}}_{HB} + E_{HA}^{2} + 2 E_{HB} \cdot E_{HA} \cdot c o s 2 θ}$

$= \sqrt{E_{HA}^{2} + E_{HA}^{2} + 2 E_{HA} + c o s 2 θ} (E_{HB} = E_{HA})$

$= \sqrt{2 E_{HA}^{2} + E_{HA}^{2} c o s 2 θ} = \sqrt{2 E_{HA}^{2} (1 + c o s 2 θ)}$

$= \sqrt{2 E_{hA}^{2} \cdot 2 c o s^{2} θ} [1 + c o s 2 θ = 2 c o s^{2} θ]$

$= 2 E_{HA} \cdot c o s θ = 2 \cdot [\frac{1}{4 π ϵ_{0}} \cdot \frac{q}{r^{2} + l^{2}}] \cdot \frac{1}{\sqrt{r^{2} + l^{2}}}$

$= 2 \cdot \frac{1}{4 π ε_{0}} \cdot \frac{ql}{{(r^{2} + l^{2})}^{3 / 2}} = \frac{1}{4 π ε_{0}} \cdot \frac{P}{{(r^{2} + l^{2})}^{3 / 2}}$

In case of short dipole $l < {r i.e., l}^{2} < < < r^{2}$

hence $l^{2}$ can be neglected.

$\Rightarrow E_{net} = \frac{1}{4 π ϵ_{0}} \cdot \frac{P}{r^{3}}$

If $α$ is the angle formed by ${\vec{E}}_{n e t}$ with ${\vec{E}}_{HA}$ , then

$t a n α = \frac{E_{HB} \cdot sin 2 θ}{E_{HA} + E_{HB} cos 2 θ} = \frac{E_{HA} \cdot sin 2 θ}{E_{HA} (1 + cos 2 θ)} = \frac{2 sin θ \cdot cos θ}{2 {cos}^{2} θ}$

$\Rightarrow t a n α = t a n θ \Rightarrow α = θ$

With the help of a labelled diagram, obtain an expression for the electric field intensity E at any point P on the equatorial line (broad-side-on position) of an electric dipole.

Important Questions on Electric Charges and Fields

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