An electric dipole is placed as shown in the figure.
The electric potential
(in $10^{2} \mathrm{~V}$

Question

An electric dipole is placed as shown in the figure.
The electric potential
(in $10^{2} \mathrm{~V}$ ) at point $P$ due to the dipole is $\left(\varepsilon_{0}=\right.$ permittivity of free space and $\left.\frac{1}{4 \pi \varepsilon_{0}}=K\right)$

Professor PC Thomas Classes · Answer

Here, the dipole distance is nearly equal to distance between the point $P$ thus we cannot apply the dipole formula direction. We have to calculate potential Hence, electric potential due to charge $q$ is given by,
due to individual charges
Potential due to charge ' $q$ ' at point ' $P$ ' is
$$
\begin{align*}
& V_{q}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}=K q \frac{1}{(R) 	imes 10^{-2}} \
& =\frac{K q\left(10^{2}ight)}{2} 	ag{i}
\end{align*}
$$

Potential due to charge ' $-q$ ' at point ' $p$ ' is,
$$
\begin{equation*}
V_{-q}=\frac{1}{4 \pi \varepsilon_{0}} \frac{(-q)}{\left(8 	imes 10^{-2}ight)}=\frac{-k q\left(10^{2}ight)}{8} 	ag{ii}
\end{equation*}
$$

From equation (i) and (ii), we get
$$
\begin{aligned}
& V_{	ext {net }}=k q\left(10^{2}ight)\left(\frac{1}{2}-\frac{1}{8}ight) \
& V_{	ext {net }}=\left(\frac{3}{8}ight) q k\left(10^{2}ight) \mathrm{V}
\end{aligned}
$$