Question

The potential at a point P due to an electric dipole is $1.8\times {10}^{5}V$. If P is at a distance of 50 cm apart from the centre O of the dipole and if CP makes an angle ${60}^o$ with the positive side of the axial line of the dipole, what is the moment of the dipole?

• A. 10 C-m
• B. ${10}^{-3}$ C-m
• C. ${10}^{-4}$ C-m
• D. ${10}^{-5}$ C-m

Answer 1

The correct option is D ${10}^{-5}$ C-m

$V=\frac{1}{4\pi {ϵ}_o}\frac{pcos\theta }{r^2}$

$1.8\times {10}^5=9\times {10}^9\times \frac{pcos{60}^o}{(0.5{)}^2}$

$p=\frac{1.8\times {10}^5\times 0.25\times 2}{9\times {10}^9}={10}^{-5}C-m$