Question
Derive an expression for potential due to a dipole for distances large compared to the size of the dipole. How is the potential due to dipole different from that due to a single charge?
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Answer
Consider origin at the centre of dipole. As per superposition principle, potential due to dipole will be the sum of potentials due to charges $q$ and $-q$
$V=\frac{1}{4 \pi \varepsilon_0}\left[\frac{q}{r_1}-\frac{q}{r_2}\right]$
Where,
$r_1$ and $r_2=$ distances of point $P$ from $q$ and $-q$.
$ r_1^2=r^2+a^2-2 \operatorname{arcos} \theta $
$ r_2^2=r^2+a^2+2 \operatorname{arcos} \theta $
If r is greater than a, and taking terms upto first order in $a / r$
$ r _1^2= r ^2\left[1-\frac{2 a \cos \theta}{r}+\frac{a^2}{r^2}\right]$
$=r^2\left[1-\frac{2 a \cos \theta}{r}\right]$
Also, $r _2^2= r ^2\left[1+\frac{2 a \cos \theta}{r}\right]$
With the help of Binomial theorem, keeping terms upto first order is shown below:
$\frac{1}{r_1} \equiv \frac{1}{r}\left[1-\frac{2 a \cos \theta}{r}\right]^{\frac{1}{2}}$
$\equiv \frac{1}{r}\left[1+\frac{a}{r} \cos \theta\right]$
$\frac{1}{r_2} \equiv \frac{1}{r}\left[1+\frac{2 a \cos \theta}{r}\right]^{\frac{1}{2}}$
$\equiv \frac{1}{r}\left[1-\frac{a}{r} \cos \theta\right] $
As $p = qa$
$ V=\frac{1}{4 \pi \varepsilon_0} \frac{q(2 a) \cos \theta}{r^2}$
$V=\frac{p}{4 \pi \varepsilon_0} \cdot \frac{\cos \theta}{r^2} $
Now, $p \cos \theta=\vec{p} . \hat{r}$
Hence electric potential of dipole for distances large compared to size of dipole is given as below :
$ V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\vec{p} \hat{r}}{r^2} \text { for } r \gg a$
For potential at any point on axis, $\theta=[0, \pi]$
$V= \pm \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} $
potential is positive when $\theta=0$
potential is negative when $\theta=\pi$
Hence, electric potential falls at large distance, as $\frac{1}{r^2}$ and not as $\frac{1}{r}$
$V=\frac{1}{4 \pi \varepsilon_0}\left[\frac{q}{r_1}-\frac{q}{r_2}\right]$
Where,
$r_1$ and $r_2=$ distances of point $P$ from $q$ and $-q$.
$ r_1^2=r^2+a^2-2 \operatorname{arcos} \theta $
$ r_2^2=r^2+a^2+2 \operatorname{arcos} \theta $
If r is greater than a, and taking terms upto first order in $a / r$
$ r _1^2= r ^2\left[1-\frac{2 a \cos \theta}{r}+\frac{a^2}{r^2}\right]$
$=r^2\left[1-\frac{2 a \cos \theta}{r}\right]$
Also, $r _2^2= r ^2\left[1+\frac{2 a \cos \theta}{r}\right]$
With the help of Binomial theorem, keeping terms upto first order is shown below:
$\frac{1}{r_1} \equiv \frac{1}{r}\left[1-\frac{2 a \cos \theta}{r}\right]^{\frac{1}{2}}$
$\equiv \frac{1}{r}\left[1+\frac{a}{r} \cos \theta\right]$
$\frac{1}{r_2} \equiv \frac{1}{r}\left[1+\frac{2 a \cos \theta}{r}\right]^{\frac{1}{2}}$
$\equiv \frac{1}{r}\left[1-\frac{a}{r} \cos \theta\right] $
As $p = qa$
$ V=\frac{1}{4 \pi \varepsilon_0} \frac{q(2 a) \cos \theta}{r^2}$
$V=\frac{p}{4 \pi \varepsilon_0} \cdot \frac{\cos \theta}{r^2} $
Now, $p \cos \theta=\vec{p} . \hat{r}$
Hence electric potential of dipole for distances large compared to size of dipole is given as below :
$ V=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\vec{p} \hat{r}}{r^2} \text { for } r \gg a$
For potential at any point on axis, $\theta=[0, \pi]$
$V= \pm \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} $
potential is positive when $\theta=0$
potential is negative when $\theta=\pi$
Hence, electric potential falls at large distance, as $\frac{1}{r^2}$ and not as $\frac{1}{r}$
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