1. Tardigrade
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  3. Physics
  4. The potential at a point P due to an electric dipole is 1.8 × 105 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° with the positive side of the axial line of the dipole, what is the moment of the dipole?

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Q. 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° with the positive side of the axial line of the dipole, what is the moment of the dipole?

Electric Charges and Fields

Solution:

$V = \frac{1}{4\pi\varepsilon_{0}} \frac{ p \cos\theta}{r^{2}}$
Here, $ V = 1.8 \times10^{5} V , \theta =60^{\circ} $
$r =50 \times10^{-2} = 0.5 m$
$ \therefore 1.8 \times10^{5} = 9 \times10^{9} \times\frac{p\cos60^{\circ}}{\left(0.5\right)^{2}} $
or $p = \frac{1.8 \times10^{5} \times0.25 \times2}{9 \times10^{9}} = 10^{-5} C - m $