A short electric dipole has a dipole moment of $16 \times 10^{- 9} Cm$ . The electric potential due to the dipole at a point at a distance of 0.6m from the centre of the dipole, situated on a line making an angle of $60^{0}$ with the dipole axis is: $(\frac{1}{4 π \in_{0}} = 9 \times 10^{9} {Nm}^{2} / C^{2})$

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

200 V

400 V

50 V

Zero

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Electric Potential at an angle of $60^{0}$ is
$V = \frac{K P \cos θ}{r^{2}}$
$There k = \frac{1}{4 π ϵ_{0}} = 9 \times 10^{9} {Nm}^{2} / C^{2}$
$P = 16 \times 10^{9} C m$
r = 0.6m
$∴ V = \frac{9 \times 10^{9} \times 16 \times 10^{9} \times \cos 60^{0}}{{(0.6)}^{2}}$
$V = \frac{9 \times 16 \times 1}{0.6 \times 0.6 \times 2}$
$V = 200 V o l$