# (a) Draw the equipotential surfaces corresponding to a uniform electric field in the z-direction.(b) Derive an expression for the electric potential at any point along the axial line of an electric dipole.

We know that the electric field strength is proportional to the distance between the surface, so, the surface will be evenly spaced.

Such that, the electric field is always perpendicular to the equipotential surface.

Hence, the equipotential surfaces are going to be infinite plane sheets in the z-direction parallel to the x-y plane.

(b) $\Rightarrow$ Let us consider an electric dipole having $-q$ and $q$ charges separated by a certain distance $(2a)$.

Now, we need to find the electric potential at any point along the axial line of an electric dipole.

Let the point $'P'$ along the axial line whose distance from the centre of the dipole is $'r'$

Thus, the electric potential at $'p'$ due to charge $A$ will be $V_{1}$ and the electric potential at $'p'$ due to charge $B$ will be $V_{2}$

Now, calculating the net potential at point $'p'$

We have,

$V_{net}=V_{1}+V_{2}$            $\left [ \because v=\frac{1}{4\pi \epsilon _{0}}\frac{q}{r} \right ]$

$V_{net}=\frac{-1}{4\pi \epsilon _{0}}\frac{q}{(r+a)}+\frac{1}{4\pi \epsilon _{0}}\frac{q}{(r-a)}$

$V_{net}=\frac{q}{4\pi \epsilon _{0}}\left [ \frac{1}{r-a}-\frac{1}{r+a} \right ]$

$V_{net}=\frac{1}{4\pi \epsilon _{0}}q\left [ \frac{r+a-r+a}{r^{2}+a^{2}} \right ]$

$V_{net}=\frac{1}{4\pi \epsilon _{0}}\frac{q(2a)}{r^{2}-a^{2}}$

We know that, dipole moment  $\vec{p}=2qa$

$V_{net}=\frac{1}{4\pi \epsilon _{0}}\frac{p}{(r^{2}-a^{2})}$

For $r> > a$

$V_{net}=\frac{1}{4\pi \epsilon _{0}}\frac{p}{r^{2}}$

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