Question

The value of electric potential at any point due to any electric dipole is:-
A. $$k{\displaystyle \frac{\overrightarrow{P}\times \overrightarrow{r}}{r^2}}$$
B. $$k{\displaystyle \frac{\overrightarrow{P}\times \overrightarrow{r}}{r^3}}$$
C. $$k{\displaystyle \frac{\overrightarrow{P}\cdot \overrightarrow{r}}{r^2}}$$
D. $$k{\displaystyle \frac{\overrightarrow{P}\cdot \overrightarrow{r}}{r^3}}$$

Answer 1

Hint:The electric dipole is a system of two charges that are separated by a finite distance. The dipole has a property called dipole moment which is equal to the product of the charges and the distance of separation between them. To find the value of electric potential at any point due to any electric dipole we use electric potential due to both charges.

Formula used: >
The potential due to this dipole is given as
$$V={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{p\cos \theta }{r^2}}$$
Where p is dipole moment, r is distance from dipole, k is coulomb’s constant,$$k={\displaystyle \frac{1}{4\pi {\epsilon }_0}}=9\times {10}^{9}N{m}^2$$ and $${\epsilon }_0$$ is permittivity.
Dipole moment is given as,
$$p=q\times 2l$$
Where q is the charge and 2l is the distance of two charges.

Complete step by step solution:

Image: A point P is at some distance r from the dipole.

In figure AB the dipole of distance between them is 2l. Point P is at a distance r from the midpoint of AB.
BP=CP=OP-OC= $$r-l\cos \theta$$
AP=DP=OP-OD=$$r+l\cos \theta$$
Electric potential due to -q,
$$V_1={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{(-q)}{AP}}$$
Electric potential due to +q,
$$V_2={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{(+q)}{BP}}$$

Now electric potential at P can be:
$$V=V_1+V_2$$
$$\Rightarrow V=qk\left[-{\displaystyle \frac{1}{r+l\cos \theta }}+{\displaystyle \frac{1}{r-l\cos \theta }}\right]$$ where $$k={\displaystyle \frac{1}{4\pi {\epsilon }_0}}$$
$$\Rightarrow V={\displaystyle \frac{qk2l\cos \theta }{r^2-l^2{\cos }^2\theta }}$$
As distance between the dipoles are less, so we can write;
$$l^2{\cos }^2\theta \approx 0$$
Also we know that $$p=q\times 2l$$

By using these values in above, we get
$$V=k{\displaystyle \frac{pr\cos \theta }{r^3}}$$
$$\therefore V=k{\displaystyle \frac{\overrightarrow{p}\cdot \overrightarrow{r}}{r^3}}$$
Therefore the value of electric potential at any point due to any electric dipole is $$V=k{\displaystyle \frac{\overrightarrow{p}\cdot \overrightarrow{r}}{r^3}}$$

Hence option D is the correct answer.

Note: As we know electric dipoles consist of two charges equal in magnitude (q) but opposite in nature one is a positive charge and other is a negative charge. These two charges are separated by a distance. Electric potential obeys superposition principle due to electric dipoles as a whole can be sum of potential due to both the charges positive and negative.