Derive equation electric potential of dipole
Answers
Answer:
Therefore, the electric potential due to an electric dipole at a given point is equal to KPcosθr2−a2cos2θ. ... Therefore, V=KPcos(π2)r2−a2cos2(π2). We know cos(π2)=0. Hence, V=KPcos(π2)r2−a2cos2(π2)=KP(0)r2−a2(0)=0.
Explanation:
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=1V=
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.