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Q. 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 :
$(\frac {1}{4\pi \epsilon_o}= 9 \times 10^9 N\,m^2/C^2)$

NEETNEET 2020Electrostatic Potential and Capacitance

Solution:

The electric potential due to the dipole,
$V=\frac{kp\,cos\,theta}{r^{2}}$
$=\frac{9\times 10^{9}\times 16\times 10^{-9}}{(0.6)^{2}}\times \frac{1}{2}$
$V=200\,V$