A short electric dipole has a dipole moment of 16 × 10⁻⁹ Cm. The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of 60° with the dipole axis is :

\(\left(\frac{1}{4 \pi \epsilon_{0}} = 9 \times 10^{9} \text{Nm}^{2}/\text{C}^{2}\right)\)

Question

Question

This question was previously asked in

NEET 2020 Official Paper (Held On: 13 September, 2020)

Answer 1

CONCEPT:

When an electric dipole is having dipole moment then its electric potential "V" at a point of a distance "r" is written as:

\(V = \frac{{kp\,\cos θ }}{{{r^2}}}\) ---- (1)

CALCULATION:

Given:

p = 16 × 10−9 Cm, r = 0.6 m, θ = 60°

Using equation (1) we have ;

\(V = \frac{{kp\,\cos θ }}{{{r^2}}}\)

⇒\(V = \frac{{9 \times {{10}^9} \times 16 \times {{10}^{ - 9}} \times \cos \,60}}{{0.36}}\)

⇒V = 200 V

Hence, option 4) is the correct answer.