5. Derive an expression for the potential due to an electric dipole at any point (r, o). How is this expression modified when the point lies on the equatorial line of the dipole?
Answers
Answer:
Let an electric dipole consist of two equal and opposite point charges –q at A and +q at B ,separated by a small distance AB =2a ,with centre at O.
The dipole moment, p=q×2a
We will calculate potential at any point P, where
OP=r and ∠BOP=θ
Let BP=r
1
and AP=r
2
Draw AC perpendicular PQ and BD perpendicular PO
In ΔAOC cosθ=OC/OA=OC/a
OC=acosθ
Similarly, OD=acosθ
Potential at P due to +q=
4πϵ
0
1
r
2
q
And Potential at P due to −q=
4πϵ
0
1
r
1
q
Net potential at P due to the dipole
V=
4πϵ
0
1
(
r
2
q
−
r
1
q
)
⟹V=
4πϵ
0
q
(
r
2
1
−
r
1
1
)
Now, r
1
=AP=CP
=OP+OC
=r+acosθ
And r
2
=BP=DP
=OP–OD
=r−acosθ
V=
4πϵ
0
q
(
r−acosθ
1
−
r+acosθ
1
)
=
4πϵ
0
q
(
r
2
−a
2
cos
2
θ
2acosθ
)
=
r
2
−a
2
cos
2
θ
pcosθ
(Since p=2aq)
Special cases:-
(i) When the point P lies on the axial line of the dipole, θ=0
∘
cosθ=1
V=
r
2
−a
2
p
If a<<r, V=
r
2
p
Thus due to an electric dipole ,potential, V∝
r
2
1
(ii) When the point P lies on the equatorial line of the dipole, θ=90
∘
cosθ=0
i.e electric potential due to an electric dipole is zero at every point on the equatorial line of the dipole.
Explanation:
this is the answer