Question: Q. 4. (i) Define electric dipole moment. Is it a scalar or a vector quantity? Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.
(ii) Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero.
R [O.D. I, U, III 2013]
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Solution:
Ans. (i) Electric dipole moment : The strength of an electric dipole is measured by the quantity of electric dipole moment. Its magnitude is equal to the product of the magnitude of either charge and the distance between the two charges.
Electric dipole moment,
It is a vector quantity.
In vector form, it is written as
Electric field of dipole at points on the equatorial plane :
The magnitudes of the electric field due to the two charges
$$ \begin{aligned} & \left|\vec{E}{+q}\right|=\frac{q}{4 \pi \varepsilon{0}} \cdot \frac{1}{\left(r^{2}+a^{2}\right)} \ & \left|\vec{E}{-q}\right|=\frac{q}{4 \pi \varepsilon{0}} \cdot \frac{1}{\left(r^{2}+a^{2}\right)} \ & \left|\vec{E}{+q}\right|=\left|\vec{E}{-q}\right| \end{aligned} $$
The direction of $\left|\vec{E}{+q}\right|
[Negative sign shows that field is opposite to
At large distances
$$ \begin{align*} \vec{E} & =\frac{-2 q a}{4 \pi \varepsilon_{0} r^{3}} \hat{p} \tag{ii}\ \left|\vec{E}{+q}\right| & =\frac{q}{4 \pi \varepsilon{0}} \cdot \frac{1}{\left(r^{2}+a^{2}\right)} \tag{iii}\ \left|\vec{E}{-q}\right| & =\frac{q}{4 \pi \varepsilon{0}} \cdot \frac{1}{\left(r^{2}+a^{2}\right)} \tag{iv}\ \therefore \quad\left|\vec{E}{+q}\right| & =\left|\vec{E}{-q}\right| \ \because \quad \vec{p} & =2 q \hat{a} \end{align*} $$
(ii) Equipotential surface due to electric dipole :
The potential due to the dipole is zero at the line bisecting the dipole length.
[CBSE Marking Scheme, 2013]