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?
AIIMS 2010, Medium
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$V=\frac{1}{4 \pi \varepsilon_{0}} \frac{p \cos \theta}{r^{2}}$
Here, $\mathrm{V}=1.8 \times 10^{5}\, \mathrm{V}, \theta=60^{\circ}$
$\mathrm{r}=50 \times 10^{-2}=0.5\, \mathrm{m}$
$\therefore 1.8 \times 10^{5}=9 \times 10^{9} \times \frac{p \cos 60^{\circ}}{(0.5)^{2}}$
or $ p=\frac{1.8 \times 10^{5} \times 0.25 \times 2}{9 \times 10^{9}}=10^{-5}\, \mathrm{C}-\mathrm{m}$
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