NEET 2020 Physics Question 170 Solution

For an electric dipole of moment $$p = 2aq$$, when distance $$r \gg 2a$$, the electrostatic potential at coordinates $$(r, \theta)$$ relative to the dipole axis is given by: $$V = \frac{1}{4\pi\varepsilon_0} \frac{p \cos\theta}{r^2}$$.

- $$p = 16 \times 10^{-9}\text{ C m}$$ - $$r = 0.6\text{ m}$$ - $$\theta = 60^\circ$$ - $$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N m}^2\text{C}^{-2}$$ - $$\cos(60^\circ) = 0.5$$.

$$V = (9 \times 10^9) \times \frac{(16 \times 10^{-9}) \times 0.5}{(0.6)^2}$$ Notice that $$10^9 \times 10^{-9} = 1$$: $$V = \frac{9 \times 8}{0.36} = \frac{72}{0.36} = \frac{7200}{36} = 200\text{ V}$$.

The electric potential is $$200\text{ V}$$, matching Option 2.