Question

Two point charges $$\pm 3.2\times {10}^{-6}C$$ are separated by a distance of $2.4\times {10}^{-10}$ from each other. Find the electric potential due to this dipole at a point P which is at a distance $3cm$ from the center of the dipole. The position vector of P w.r.t. the center of the dipole makes an angle of ${30}^{\circ }$ with the dipole.
A. $7.68\times {10}^{-3}V$
B. $\sqrt{3}\times 3.84\times {10}^{-3}V$
C. $\sqrt{3}\times 1.97\times {10}^{-3}V$
D. $3.84\times {10}^{-3}V$

Answer 1

Hint: An electric potential is defined as the amount of work needed to move a point charge from any point to a given point without producing any acceleration. Here, we will use the electrostatic formula in which we have used the dipole moment formula. Here, we will calculate the electric potential at point P.

Formula Used:
The formula of the electric potential is given by
$V={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{p\cos \theta }{r^2}}$
Here, $V$ is the electric potential, $p$ is the dipole moment, $r$ is the distance of the point P from the center of the dipole and $\theta$ is the angle by the point P from the center of the dipole.

The formula of the dipole moment is given by
$p=q\times d$
Here, $p$ is the dipole moment, $q$ is the charge of the dipole and $d$ is the distance between the charges.

Complete step by step answer:
Here, we are considering two point charges that are separated by a distance of $2.4\times {10}^{-10}$ from each other and they are forming a dipole. Now consider a point P which is at a distance of $3cm$ from the center of dipole.
Now, the charge on the dipole, $q=\pm 3.2\times {10}^{-6}C$
Also, the distance between the two charge or the length of the dipole, $d=2.4\times {10}^{-10}m$
And, the distance of point P from the center of the dipole, $r=3cm=0.03m$
Now, the formula of the electrostatic potential is given by
$V={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{p\cos \theta }{r^2}}$
Now, we know that the formula of the dipole is given by
$p=q\times d$
Putting the above value in the above equation, we get
$V={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{q\times d\cos \theta }{r^2}}$
$\Rightarrow V=\left(9\times {10}^9\right)\times {\displaystyle \frac{3.2\times {10}^{-6}\times 2.4\times {10}^{-10}\times \cos {30}^{\circ }}{{\left(0.03\right)}^2}}$
$\Rightarrow V=3.2\times 2.4\times {10}^{-3}\times {\displaystyle \frac{\sqrt{3}}{2}}$
$\therefore V=\sqrt{3}\times 3.84\times {10}^{-3}V$
Therefore, the electrostatic potential at a point P that is at a distance $3cm$ from the center of the dipole is $\sqrt{3}\times 3.84\times {10}^{-3}V$ .

Hence, option B is the correct option.

Note:Now, you might think that dipole consists of two charges but we have taken only one charge. This is because we always get the same answer whether we take positive charge or we take negative charge. That is why, in the above solution we have used only positive value of charge.