Question

A small electric dipole is of dipole moment p. The electric potential at a distance ‘r’ from the center and making an angle $\theta$ with the axis of dipole will be:

Answer 1

Hint: Electric dipole is a couple of opposite charges given by q and -q. they are separated by the distance d. In this given space the direction of the dipole is always from -q to positive charge q. In the question we need to find potential at distance r from center. It is making an angle $\theta$ with the axis.

Formula Used:
 $$\begin{gathered} V={\displaystyle \frac{q}{4\pi {\epsilon }_{0}R}} \\ \cos \theta ={\displaystyle \frac{y^2+z^2-x^2}{2yz}} \\ (1+x^n)=1+nxforx<<1 \end{gathered}$$

Complete step-by-step solution:

First of all we have the dipole with charges q and -q at the ends.
The potential at the distance ‘r’ can be found out as follows:
First from the figure we will get,
$$\begin{gathered} USING, \\ \cos \theta ={\displaystyle \frac{y^2+z^2-x^2}{2yz}} \end{gathered}$$
Cosine formula for any triangle.
Here, z is hypotenuse.
According to our dimensions
$$\begin{gathered} \cos \theta ={\displaystyle \frac{r^2+a^2-{r_1}^2}{2ra}} \\ \Rightarrow {r_1}^2=r^2+a^2-2ra\cos \theta \end{gathered}$$
By taking square root on both sides we get,
$$\begin{gathered} r_1=(r^2+a^2-2ra\cos \theta {)}^{{\displaystyle \frac{1}{2}}} \\ taking{}^{\prime }{r}^{\prime }common \\ \Rightarrow r_1=r{\left[1-{\displaystyle \frac{a}{r}}\cos \theta \right]}^{{\displaystyle \frac{1}{2}}} \end{gathered}$$
Now using the identity,
$(1+x^n)=1+nxforx<<1$
We get,
$$\begin{gathered} r_1=r\left[1-{\displaystyle \frac{a}{r}}\cos \theta \right] \\ thisgives \\ \Rightarrow r_1=r-a\cos \theta \\ \Rightarrow r_2=r+a\cos \theta \end{gathered}$$
We know for the potential we need to find individual potentials from each charge and add them. So,
$$\begin{gathered} V={\displaystyle \frac{q}{4\pi {\epsilon }_{0}R}} \\ addingall \\ \Rightarrow V={\displaystyle \frac{q}{4\pi {\epsilon }_0{r}_1}}+{\displaystyle \frac{q}{4\pi {\epsilon }_0{r}_2}} \end{gathered}$$
Putting the values taken found out above and solving,
$$\begin{gathered} \Rightarrow {\displaystyle \frac{q}{4\pi {\epsilon }_0}}\left[{\displaystyle \frac{1}{r-a\cos \theta }}-{\displaystyle \frac{1}{r+a\cos \theta }}\right] \\ \Rightarrow {\displaystyle \frac{q}{4\pi {\epsilon }_0}}\left[{\displaystyle \frac{r+a\cos \theta -r+a\cos \theta }{r^2-a^2{\cos }^2\theta }}\right] \end{gathered}$$
Further solving the equations and neglecting a as it is very small we get
$$\begin{gathered} Asp=q(2a) \\ \Rightarrow V={\displaystyle \frac{p\cos \theta }{4\pi {\epsilon }_0(r^2-a^2{\cos }^2\theta )}} \\ \Rightarrow V={\displaystyle \frac{p\cos \theta }{4\pi {\epsilon }_0(r^2)}}.....(ifaissmall) \end{gathered}$$

Note: Potential at the point is always an addition of potential due to all charges/dipoles. Potential negative means whatever the system is under study it is stable. If potential is positive it is an unstable system. Force varies inversely with square of the distance and potential inversely with first power of distance.