Question

Define electric dipole. Derive an expression for an electric field intensity due to an electric dipole at a point on the equatorial plane of the electric diagram. Draw necessary diagram.

Accepted Answer

The electric dipole is a pair of two equal and opposite charges, $+ q$ and $- q$ are separated by a very small distance. The algebraic sum of the charge on the dipole is zero and the electric field of the dipole is not zero.

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

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

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

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

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

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

Now, according to the 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.

$∴ E = \frac{P}{4 π ε_{0} x^{3}}$

Here, $P = 2 q a =$ Dipole moment

Therefore, electric field intensity due to an electric dipole at a point on the equatorial plane is $E = \frac{P}{4 π ε_{0} x^{3}}$

Define electric dipole. Derive an expression for an electric field intensity due to an electric dipole at a point on the equatorial plane of the electric diagram. Draw necessary diagram.

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