Two charges +q and -q are placed at points A(a,0) and B(-a,0) on the x-axis. The potential due to this system of charges on point P(r, θ) with position vector \(\vec r\)will be? Given that q, a are positive constants and r >> a.

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

• Potential due to a dipole with dipole moment \(\vec p\) at a point P(r,θ) having position vector \(\vec r\) with respect to the dipole moment orientation is given by (if r >> a):

\(⇒ V=\frac{⃗ p.⃗ r}{4\piϵ_{0}r^3}\)

EXPLANATION:

• Given:
  • The dipole is placed at the origin

• The dipole moment is

\(\Rightarrow \vec p = q \times 2a \hat i = 2qa \hat i\)​

• The position vector of point P is \(\vec r\)
•  Potential due to a dipole with dipole moment \(\vec p\) at a point P with position vector \(\vec r\) with respect to the dipole moment orientation is given by (if r >> a)

\(⇒ V=\frac{\vec p.\vec r}{4\piϵ_{0}r^3} = \frac{2a\hat i.\vec r}{4\piϵ_{0}r^3} = \frac{a\hat i.\vec r}{2\piϵ_{0}r^3}\)

Therefore option 2 is correct.