Q 12-02-018NEETNEET 2020Top questionMedium
A short electric dipole has a dipole moment of $16 \times 10^{-9}$ C m. 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 : $\left(\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^9\ \text{N m}^2/\text{C}^2\right)$
Answer: (B) $200$ V
$$V = \frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^2} = \frac{9 \times 10^9 \times 16 \times 10^{-9} \times 0.5}{(0.6)^2} = \frac{72}{0.36} = 200\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics