Question

a short electric dipole has dipole moment 1 ×10^9 cm. determine the electric potential due to the dipole at a point distance 0.3 m from the centre of the dipole situated 1)on the axis line 2) on the equilibrial line 3)on a line making an angle of 60 with the dipole axis​

Answer 1

Explanation:

hope this helps you

mark as brainslist

Answer 2

Answer:

Part 1) Axial line potential is 10V.

Part 2) equatorial line potential is 0V.

Part 3)line making an angle α=60 with the dipole axis has potential of 5V.

Explanation:

For an ideal dipole, the potential at any general point at a distance r from it making an angle α with dipole axis can be given the formula,

$V=\frac{KPcos\alpha }{r^{2} }$

where K=$\frac{1}{4\pi E_{0} } =9*10^{9}$

P= dipole moment of the given electric dipole= $10^{-9} Cm$

Part1:

For the axis line α=0°

V=  $\frac{K*10^{-9}*cos0 }{0.3^{2} }$

On solving V=10V

Hence, On the Axial line potential is 10V.

Part2:

For the equatorial line α=90°

$V=\frac{K*10^{-9}*cos90 }{0.3^{2} }$

V=0V

Hence, On the equatorial line potential is 0V.

Part3:

For a line making an angle α=60 with the dipole axis.

$V=\frac{K*10^{-9}*cos60 }{0.3^{2} }$

On solving V=5V

Hence, On the line making an angle α=60 with the dipole axis potential is 5V.

​