Question

A short electric dipole has a dipole moment of
16x 10-9 Cm. The electric potential due to the
dipole at a point at a distance of 0.6 m from the
centre of the dipole, situated on a line making an
angle of 60° with the dipole axis is:
1 N
=9x 109 Nm²/C2
4TTEO​

Answer 1

Answer :

Dipole moment = 16 × 10$^{-9}$ Cm

We have to find electric potential due to the dipole at a distance of 0.6m from the centre of the dipole situated on a line making an angle 60°$.$

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◈ Electric potential at distance r from centre of dipole making an angle Φ with dipole moment is given by

$\dag\:\boxed{\bf{V=\dfrac{KP\cos\phi}{r^2}}}$

$\leadsto\sf\:V=\dfrac{(9\times 10^9)(16\times 10^{-9})\cos60\degree}{(0.6)^2}$

$\leadsto\sf\:V=\dfrac{144\times 0.5}{0.36}$

$\leadsto\sf\:V=\dfrac{72}{0.36}$

$\leadsto\:\boxed{\bf{V=200\:volt}}$

Answer 2

Answer:

Dipole moment = 16 × 10^{-9}

−9

Cm

We have to find electric potential due to the dipole at a distance of 0.6m from the centre of the dipole situated on a line making an angle 60°..

_________________________________

◈ Electric potential at distance r from centre of dipole making an angle Φ with dipole moment is given by

\dag\:\boxed{\bf{V=\dfrac{KP\cos\phi}{r^2}}}†

V=

r

2

KPcosϕ

\leadsto\sf\:V=\dfrac{(9\times 10^9)(16\times 10^{-9})\cos60\degree}{(0.6)^2}⇝V=

(0.6)

2

(9×10

9

)(16×10

−9

)cos60°

\leadsto\sf\:V=\dfrac{144\times 0.5}{0.36}⇝V=

0.36

144×0.5

\leadsto\sf\:V=\dfrac{72}{0.36}⇝V=

0.36

72

\leadsto\:\boxed{\bf{V=200\:volt}}⇝

V=200volt