Question

Draw a circuit diagram to find the ratio of EMFs of two primary cells using potentiometer using sum and difference method. Hence derive the necessary expression.​

Answer 1

Explanation:

Consider an electric dipole AB in which point charges +q and -q are separated by a distance 2l. Let P be a point on it's axis seperated by a distance r from the dipole's center O.

Electric dipole moment, p = q (2l)

Electric potential (V

A

) due to +q charge

V

A

=

4πϵ

0

1

r−l

q

Electric potential (V

B

) due to +q charge

V

B

=

4πϵ

0

1

r+l

−q

Therefore, total electric potential at P due to dipole will be

V=

4πϵ

0

q

[d

r−l

1

−

r+l

1

]

=

4πϵ

0

q

r

2

−l

2

2l

=

4πϵ

0

p

r

2

−l

2

1

Now, l<<r (short dipole)

V=

4πϵ

0

p

r

2

1

(for axial position)

For equatorial position, V=0 as charges are equal in magnitude with opposite signs they cancel each other's effect.

solution