Question

derive the expression for the electric potential at any point along the axial line of an electric dipole and it's S.I. unit and dimension?​

Answer 1

Explanation:

Let P be the axial point at distance x from the centre of the dipole electric potential V at point P is given by,

V=V

+q

+V

−q

=K{

x−a

q

x+a

−q

}

=

x

2

−a

2

Kq2a

=

(x

2

−a

2

)

KP

P: dipole moment =2aq

II

nd

Part

2) E=

r

V

⇒25=

r

10

r=

25

10

=0.4m

∴ distance of the charge from point of observation =0.4m

Now, i) V=

r

KQ

⇒10×0.4=9×10

9

×Q

∴Q=4.45×10

−10

C

∴ Magnitude of charge =4.45×10

−10

C

distance =0.4m