Electric Potential Of A Dipole And System Of Charges - Practice Questions & MCQ

Electric Potential due to an Electric Dipole at a Point on the Axial Line

As shown in the above figure We want to find out Electric Potential due to an Electric Dipole at a Point M which is on axial line and at a distance r from the center of a dipole.

Where $V_1$ and $V_2$ is the Electric Potential at M due to $-q$ and $+q$ charges respectively.

$
\begin{aligned}
& V_1=\frac{k q}{(r+a)} \\
& \begin{aligned}
& V_2=\frac{k q}{(r-a)} \\
& \begin{aligned}
V_{\text {net }} & =V_2-V_1 \\
V_{\text {net }} & =V_1+V_2 \\
& =\frac{-k q}{(r+a)}+\frac{k_q}{(r-a)} \\
& =k_q\left\{\frac{1}{r-a}-\frac{1}{r+a}\right\} \\
& =k q\left\{\frac{(r+a)-(r-a)}{(r-a)(r+a)}\right\}
\end{aligned}
\end{aligned}
\end{aligned}
$

So $V_{\text {net }}=\frac{2 k q a}{r^2-a^2}$
Using $P=q(2 a)$
So $V_{\text {net }}=\frac{k P}{r^2-a^2}$
- if $r \gg a$
then $V_{n e t}=\frac{K P}{r^2}=\frac{P}{4 \pi \epsilon_0 r^2}$

Electric potential due to an Electric Dipole at a Point on the Equitorial line.

As shown in the above figure We want to find out Electric potential due to an Electric Dipole at a Point M which is on the Equitorial line and at a distance r from the center of a dipole.

Where $V_1$ and $V_2$ is the Electric Field Intensity at M due to $-q$ and $+q$ charges respectively.

$
\begin{aligned}
& V_1=-\frac{1}{4 \pi \epsilon_0} * \frac{q}{\sqrt{r^2+a^2}} \\
& V_2=\frac{1}{4 \pi \epsilon_0} * \frac{q}{\sqrt{r^2+a^2}} \\
& V_{\text {net }}=V_2-V_1=0
\end{aligned}
$
 

Electric potential due to a dipole at any general point-

As shown in the above figure We want to find out Electric potential due to an Electric Dipole at a Point M which at a distance r from the center of a dipole and making an angle  with the axial line.

From the figure, M is at the axial line of dipole having dipole moment as $P \cos \theta$ and M is at the Equitorial line of dipole having dipole moment as $P \sin \theta$.
So $P \sin \theta$ has no contribution in electric potential at point M.
if $r>>a$
then
$V_a=\frac{1}{4 \pi \varepsilon_0} \times \frac{2 P \cos \theta}{r^2}$ and $V_{\perp}=0$
$\mathrm{So} V_{\text {net }}=V_a=\frac{K P \cos \theta}{r^2}$