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Physics Question on Electric Dipole

The potential at a point P due to an electric dipole is 1.8×105 V1.8 \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?

A

10 C - m

B

10−3 C−m10^{-3} \, C - m

C

10−4 C−m10^{-4} \, C - m

D

10−5 C−m10^{-5} \, C - m

Answer

10−5 C−m10^{-5} \, C - m

Explanation

Solution

V=14πε0pcos⁡θr2V = \frac{1}{4\pi\varepsilon_{0}} \frac{ p \cos\theta}{r^{2}} Here, V=1.8×105V,θ=60∘V = 1.8 \times10^{5} V , \theta =60^{\circ} r=50×10−2=0.5mr =50 \times10^{-2} = 0.5 m ∴1.8×105=9×109×pcos⁡60∘(0.5)2\therefore 1.8 \times10^{5} = 9 \times10^{9} \times\frac{p\cos60^{\circ}}{\left(0.5\right)^{2}} or p=1.8×105×0.25×29×109=10−5C−mp = \frac{1.8 \times10^{5} \times0.25 \times2}{9 \times10^{9}} = 10^{-5} C - m