Correct Answer - Option 2 :
\(\frac{1}{r^2}\)
CONCEPT :
- Electric potential is equal to the amount of work done per unit charge by an external force to move the charge q from infinity to a specific point in an electric field.
\(⇒ V=\frac{W}{q}\)
- Potential due to a single charged particle Q at a distance r from it is given by:
\(⇒ V=\frac{Q}{4\piϵ_{0}r}\)
Where,
ϵ0 is the permittivity of free space and has a value of 8.85 × 10-12 F/m in SI units
EXPLANATION:
- Given that r >> a, where 2a is the distance between the two charges of the dipole
- The dipole moment is
\(⇒ \vec p = (q \times 2a )\hat p = (2aq) \hat p\)
- Potential due to a dipole with dipole moment \(\vec p\) at a point P having position vector \(\vec r\) with respect to the dipole moment orientation is given by (if r >> a)
\(⇒ V=\frac{\vec p \cdot\vec r}{4\piϵ_{0}r^3} =\frac{(2a q \hat p).(r \hat r)}{4\piϵ_{0}r^3} =\frac{2qa \hat p.\hat r}{4\piϵ_{0}r^2} \)
- There the potential falls off due to a factor of \(\frac {1}{r^2}\).
Therefore option 2 is correct.
