ELECTROSTATIC POTENTIAL AND CAPACITANCE
*Work done in bringing a unit +ve charge from infinity to any point in the electrostatic field
1 Volt
One joule of work done to move 1 coulomb of charge from infinity to the point against ES force
V=kΣ Qi/r
Vp=kQ/r
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OC=acosθ
0
1
r
2
q
4πϵ
0
1
r
1
q
V=
4πϵ
0
1
r
2
q
r
1
q
⟹V=
4πϵ
0
q
r
2
1
r
1
1
Now, r
1
=OP+OC
=r+acosθ
And r
=r−acosθ
V=
4πϵ
0
q
r
2
=OP–OD
r−acosθ
1
r+acosθ
1
4πϵ
q
0
4πϵ
=
2
2
−
)
(
−
(
)
=AP=CP
2
=
2acosθ
1
r
2
(
−
)
(
−a
=BP=DP
cos
2
2
pcosθ
∘
cosθ=1
V=
r
2
2
p
r
2
p
r
2
1
In ΔAOC cosθ=OC/OA=OC/a
Similarly, OD=acosθ
Special cases:-
Potential at P due to +q=
If a<<r, V=
∘
and AP=r
cosθ=0
And Potential at P due to −q=
(i) When the point P lies on the axial line of the dipole, θ=0
Thus due to an electric dipole ,potential, V∝
(ii) When the point P lies on the equatorial line of the dipole, θ=90
i.e electric potential due to an electric dipole is zero at every point on the equatorial line of the dipole.
Net potential at P due to the dipole
θ
)
−a
cos
θ
(Since p=2aq)
Draw AC perpendicular PQ and BD perpendicular PO
−a
Let BP=r
We will calculate potential at any point P, where
OP=r and ∠BOP=θ
The dipole moment, p=q×2a
2
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.