Question

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 $$60^o$$ with the dipole axis is :
$$\left(\dfrac{1}{4\pi \in_0} = 9\times 10^9 Nm^2/C^2\right)$$

Answer 1

Correct option is A. $$200 V$$
Electric Potential at an angle of $$60^o$$ is

$$V = \dfrac{KP \cos \theta}{r^2}$$

there $$K = \dfrac{1}{4\pi \in_0} = 9 \times 10^9 Nm^2 / C^2$$

$$P = 16\times 10^{-9} Cm$$

$$r = 0.6m$$

$$\therefore V = \dfrac{9\times 10^9 \times 16 \times 10^{-9}\times \cos 60^o}{(0.6)^2}$$

$$V = \dfrac{9\times 16 \times 1}{0.6\times 0.6\times 2}$$

$$V = 200Vol$$