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

Updated On: Jul 7, 2022
  • 10 C - m
  • $10^{-3} \, C - m$
  • $10^{-4} \, C - m$
  • $10^{-5} \, C - m$
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The Correct Option is D

Solution and Explanation

$V = \frac{1}{4\pi\varepsilon_{0}} \frac{ p \cos\theta}{r^{2}}$ 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 $
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Concepts Used:

Electric Dipole

An electric dipole is a pair of equal and opposite point charges -q and q, separated by a distance of 2a. The direction from q to -q is said to be the direction in space.

p=q×2a

where,

p denotes the electric dipole moment, pointing from the negative charge to the positive charge.

Force Applied on Electric Dipole