Question

Derive an expression for the electric field at any point on the equatorial line of an electric dipole.

Accepted Answer

Consider an electric dipole consisting of charges $- q$ and $+ q$ separated by a distance $2 a$ .

Let $E_{A}$ be the electric field at point $Q$ due to charge $- q$ and $E_{B}$ be the electric field at point $Q$ due to charge $+ q$ where $Q$ is a point on the equatorial line of the dipole whose electric field is to be measured. Let $x$ be the distance between the centre of the dipole and point $Q$ .

Now, resultant electric field at point $Q$ would be,

$E_{Q} = E_{A} + E_{B}$

$E_{A} = \frac{1}{4 π ε_{0}} (\frac{q}{{(A Q)}^{2}})$

$\Rightarrow \frac{1}{4 π ε_{0}} (\frac{q}{x^{2} + a^{2}})$ ( $∵$ According to Pythagoras, $A Q = {(x^{2} + a^{2})}^{2}$ )

$E_{B} = \frac{1}{4 π ε_{0}} (\frac{q}{{(B Q)}^{2}})$

$\Rightarrow \frac{1}{4 π ε_{0}} (\frac{q}{x^{2} + a^{2}})$ ( $∵$ According to Pythagoras, $B P = {(x^{2} + a^{2})}^{2}$ )

Now, according to parallelogram law of vector addition, the resultant electric field would become,

$E_{Q} = \sqrt{{E_{A}}^{2} + {E_{B}}^{2} + 2 E_{A} E_{B} \cos 2 θ}$

Now, since $E_{A} = E_{B}$ we can say, $E_{A} = E_{B} = E'$

$E_{Q} = \sqrt{{E'}^{2} + {E'}^{2} + 2 E' E' \cos 2 θ} \Rightarrow \sqrt{2 {E'}^{2} (1 + \cos 2 θ)} \Rightarrow E' \sqrt{2 (1 + \cos 2 θ)} \Rightarrow E' \sqrt{2 (2 \cos^{2} θ)} E_{Q} = 2 E' \cos θ - - - (i)$

Now, from $△ A O Q$ , $\cos θ = \frac{a}{\sqrt{x^{2} + a^{2}}}$

Also, we know that, $E' = \frac{1}{4 π ε_{0}} (\frac{q}{x^{2} + a^{2}})$

Putting both these values in equation $(i)$ , we get,

$E_{Q} = \frac{2}{4 π ε_{0}} (\frac{q}{x^{2} + a^{2}}) (\frac{a}{\sqrt{x^{2} + a^{2}}}) E_{Q} = \frac{p}{4 π ε_{0} {(x^{2} + a^{2})}^{\frac{3}{2}}}$

Since we know that in dipoles, $x > > > 2 a$ , then $a^{2}$ can be neglected.

$∴ E = \frac{p}{4 π ε_{0} x^{\frac{3}{2}}}$

Embibe Experts Solutions for Chapter: Solved Physics Paper-2, Exercise 1: Exercise

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