Question

The electric potential at a point on the axis of an electric dipole depends on the distance ${{r}}$ of the point from the dipole as
A. $\dfrac{{{1}}}{{{r}}} $
B. $\dfrac{{{1}}}{{{{{r}}^{{2}}}}} $
C. $r$
D. $\dfrac{{{1}}}{{{{{r}}^{{3}}}}} $

Answer 1

Hint: An electric dipole consists of a couple of opposite charges ${{q}}$ and ${{ - q}}$ separated by a distance ${{r}}$. The formula for electric potential due to dipole is given by
${{V = }}\dfrac{{{{kpcos\theta }}}}{{{{{r}}^{{2}}}}}$. For the dependency of electric potential (V) with distance (r) can be clearly seen by the given formula.

Complete step by step answer:
Formula for electric potential due to dipole is given by
$
  {{V = }}\dfrac{{{{kpcos\theta }}}}{{{{{r}}^{{2}}}}} \\
   \Rightarrow {{V }}\alpha {{ }}\dfrac{{{1}}}{{{{{r}}^{{2}}}}} \\
$
Thus, the electric dipole varies inversely to the square of a distance ${{(r)}}$.
The electric potential due to an electric dipole varies inversely to the square of the distance i.e. $\dfrac{{{1}}}{{{{{r}}^{{2}}}}}$ and the potential due to a single point charge varies inversely to the distance i.e. $\dfrac{{{1}}}{{{r}}}$. Thus, the potential due to the dipole falls faster than that due to point charges. As the distance increases from electric dipole, the effects of positive and negative charges cancel out each other.

Hence, the correct answer is option (B).

Note: The potential due to a point charge is spherically symmetric because it depends only on the distance ${{r}}$. However, the potential due to a dipole is not spherically symmetric as the potential depends on the angle between the dipole and the position vector of the point. Electric Dipoles are usually found in molecular structures caused by non-uniform charge distribution of protons and electrons. Electric dipoles are used to find the polarity of a system and knowing the polarity of the system is useful in the study of many chemical phenomena such as the normal force (the reason we don't fall through objects), surface tension and solubility.