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A short electric dipole has a dipole moment of $16 \times 10^{- 9} Cm$ . The electric potential due to the dipole at a point at a distance of 1.2 m from the centre of the dipole, situated on a line making an angle of $60^{o}$ with the dipole axis is : $(\frac{1}{4 {πε}_{0}} = 9 \times 10^{9} N m^{2} / C^{2})$

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200 V

400 V

Zero

50 V

answer is D.

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Detailed Solution

Potential due to a dipole at any point is
given by $v = \frac{pcos θ}{4 {πϵ}_{0} r^{2}}$

Here, p = $16 \times 10^{- 9} C m$

r = 0.6 m

$\frac{1}{4 {πϵ}_{0}} = 9 \times 10^{9} {Nm}^{2} / C^{2}$

$θ = 60^{\circ}$

Substituting the values,

$V = \frac{16 \times 10^{- 9} \times \frac{1}{2} \times 9 \times 10^{9}}{144 \times 10^{- 2}} = 50 V$

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