Question 14 Marks
Derive the formula of electric potential at a point due to electric dipole.
Answer
$\rightarrow$ As shown in fig. point $P$ is given at a distance $'r\ '$ from the midpoint $'O\ '$ of electric dipole and at an angle $\theta\ ($with the electric dipole moment $\vec{p})$.
We want to find electric potential at this point $P$.
$\rightarrow$ Electric potential at point $P$ due to charge $+q$,
$V _1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r_1}$
$\rightarrow$ Electric potential at point $P$ due to charge $-q$,
$V _2 =\frac{1}{4 \pi \varepsilon_0} \frac{(-q)}{r_2}$
$ =-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r_2}$
$\rightarrow$ Total electric potential at point $P$ as per super position principle,
$V = V _1+ V _2$
$\therefore V =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r_1}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r_2}$
$\therefore V =\frac{q}{4 \pi \varepsilon_0}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$
$\rightarrow$ As shown in the figure $(a),$ position vector of point $P$ with respect to origin $O$ is $\vec{r}$.
Position vector of point $P$ with respect to $+q$ is $\vec{r}_1$.
Position vector of point $P$ with respect to $-q$ is $\vec{r}_2$.
$\rightarrow$ From figure $(b),$
$\vec{r} =\vec{a}+\vec{r}_1$
$\therefore \vec{r}_1 =\vec{r}-\vec{a}$
$\therefore r_1^2 =r^2+a^2-2 r a \cos \theta$
$(\theta$ is angle between $\vec{r}$ and $\vec{a})$
$\therefore r_1^2=r^2\left( i +\frac{a^2}{r^2}-\frac{2 a \cos \theta}{r}\right)$
$\rightarrow$ But value of $\frac{a^2}{r^2}$ is very less for $r \gg a$,
so it can be neglected from equation.
$\therefore r_1^2=r^2\left(1-\frac{2 a \cos \theta}{r}\right)$
$\therefore r_1=r\left(1-\frac{2 a \cos \theta}{r}\right)^{\frac{1}{2}}$
$\therefore \frac{1}{r_1}=\frac{1}{r}\left(1-\frac{2 a \cos \theta}{r}\right)^{-\frac{1}{2}}$
View full question & answer→$\rightarrow$ As shown in fig. point $P$ is given at a distance $'r\ '$ from the midpoint $'O\ '$ of electric dipole and at an angle $\theta\ ($with the electric dipole moment $\vec{p})$.
We want to find electric potential at this point $P$.
$\rightarrow$ Electric potential at point $P$ due to charge $+q$,
$V _1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r_1}$
$\rightarrow$ Electric potential at point $P$ due to charge $-q$,
$V _2 =\frac{1}{4 \pi \varepsilon_0} \frac{(-q)}{r_2}$
$ =-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r_2}$
$\rightarrow$ Total electric potential at point $P$ as per super position principle,
$V = V _1+ V _2$
$\therefore V =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r_1}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r_2}$
$\therefore V =\frac{q}{4 \pi \varepsilon_0}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$
$\rightarrow$ As shown in the figure $(a),$ position vector of point $P$ with respect to origin $O$ is $\vec{r}$.
Position vector of point $P$ with respect to $+q$ is $\vec{r}_1$.
Position vector of point $P$ with respect to $-q$ is $\vec{r}_2$.
$\rightarrow$ From figure $(b),$
$\vec{r} =\vec{a}+\vec{r}_1$
$\therefore \vec{r}_1 =\vec{r}-\vec{a}$
$\therefore r_1^2 =r^2+a^2-2 r a \cos \theta$
$(\theta$ is angle between $\vec{r}$ and $\vec{a})$
$\therefore r_1^2=r^2\left( i +\frac{a^2}{r^2}-\frac{2 a \cos \theta}{r}\right)$
$\rightarrow$ But value of $\frac{a^2}{r^2}$ is very less for $r \gg a$,
so it can be neglected from equation.
$\therefore r_1^2=r^2\left(1-\frac{2 a \cos \theta}{r}\right)$
$\therefore r_1=r\left(1-\frac{2 a \cos \theta}{r}\right)^{\frac{1}{2}}$
$\therefore \frac{1}{r_1}=\frac{1}{r}\left(1-\frac{2 a \cos \theta}{r}\right)^{-\frac{1}{2}}$