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, $p=q \times 2 a$

It is a vector quantity.

In vector form, it is written as $\vec{p}=q \times 2 \vec{a} \hat{p}$, where the direction of $2 \vec{a} \hat{p}$ is from negative charge to positive charge.

Electric field of dipole at points on the equatorial plane :

The magnitudes of the electric field due to the two charges $+q$ and $-q$ are given by,

$$ \begin{aligned} & \left|\vec{E}{+q}\right|=\frac{q}{4 \pi \varepsilon{0}} \cdot \frac{1}{\left(r^{2}+a^{2}\right)} \ & \left|\vec{E}{-q}\right|=\frac{q}{4 \pi \varepsilon{0}} \cdot \frac{1}{\left(r^{2}+a^{2}\right)} \ & \left|\vec{E}{+q}\right|=\left|\vec{E}{-q}\right| \end{aligned} $$

The direction of $\left|\vec{E}{+q}\right|$ and $\left|\vec{E}{-q}\right|$ are shown in the figure. The components normal to the dipole axis cancel out. The components along the dipole axis add up.

$\therefore$ Total electric field

$$ \vec{E}=-\left(E_{+q}+E_{-q}\right) \cos \theta \hat{p} $$

[Negative sign shows that field is opposite to ${ }^{p}$ ]

$$ \begin{equation*} \vec{E}=\frac{-2 q a}{4 \pi \varepsilon_{0}\left(r^{2}+a^{2}\right)^{3 / 2}} \hat{p} \tag{i} \end{equation*} $$

At large distances $(r»a)$, this reduces to

$$ \begin{align*} \vec{E} & =\frac{-2 q a}{4 \pi \varepsilon_{0} r^{3}} \hat{p} \tag{ii}\ \left|\vec{E}{+q}\right| & =\frac{q}{4 \pi \varepsilon{0}} \cdot \frac{1}{\left(r^{2}+a^{2}\right)} \tag{iii}\ \left|\vec{E}{-q}\right| & =\frac{q}{4 \pi \varepsilon{0}} \cdot \frac{1}{\left(r^{2}+a^{2}\right)} \tag{iv}\ \therefore \quad\left|\vec{E}{+q}\right| & =\left|\vec{E}{-q}\right| \ \because \quad \vec{p} & =2 q \hat{a} \end{align*} $$

$$ \begin{equation*} \therefore \quad \vec{E}=\frac{-\vec{p}}{4 \pi \varepsilon_{0} r^{3}} \tag{3} \end{equation*} $$

(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]