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A short electric dipole has a dipole moment of 16×109Cm16 \times 10^{-9} C m. The electric potential due to the dipole at a point at a distance of 0.6m0.6 m from the centre of the dipole, situated on a line making an angle of 6060^{\circ} with the dipole axis is : (14πϵ0=9×109Nm2/C2\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 N m^2/C^2)

1

50 V

2

200 V

3

400 V

4

zero

Step-by-Step Solution

V=kpcosθr2V = \frac{kp \cos \theta}{r^2}. V=9×109×16×109×cos600.36=200VV = \frac{9 \times 10^9 \times 16 \times 10^{-9} \times \cos 60^{\circ}}{0.36} = 200 V

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