Charged Ring Potential and Field.

27 Pages·2013·0.15 MB·English

Closed Form Solutions for Potential and Field
of a Charged Ring
Kurt Nalty
July 18, 2013
Abstract
Thisnoteprovidesastepbystepcalculationforthevoltagepoten-
tial and and electric field associated with a charged filament. At the
end of the note is a listing of a quick numerical verification of closed
form and discrete sum models.
Summary of Results
The source is a ring of charge q and radius R in the XY plane at z = 0. The
observation point in cylindrical coordinates is (ρ,φ,z).
(cid:115)
4ρR
k =
(ρ+R)2 +z2
q 1
V(ρ,φ,z) = K(k)
(cid:112)
2π2(cid:15) (ρ+R)2 +z2
q (cid:18)K(k) (cid:20)E(k) z2 +R2 −ρ2 (cid:21)(cid:19)
E = −
ρ 4π(cid:15)ρ(cid:112)(R+ρ)2 +z2 π π (R−ρ)2 +z2
E = 0
φ
(cid:18) (cid:19)
q E(k) 2ρz
E =
z 4π(cid:15)ρ(cid:112)(ρ+R)2 +z2 π (R−ρ)2 +z2
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Potential of the Charged Ring
Inthissection,wewillcalculatetheelectrostaticpotentialassociatedwiththe
charged ring. Just as a mathematical point charge has an infinity associated
with the field at the charge point, we will likewise have an infinity for the
field at the ring radius as well. Traditionally, we don’t try to measure the
voltage at the ring, itself.
For these calculations, I will be using MKS units. I will put a ring of
charge q and radius R in the XY plane at z = 0. I will place my observation
(cid:112)
point in cylindrical coordinates at (ρ,φ,z), where ρ = x2 +y2. Due to
cylindrical symmetry, this same potential will be found all around the circle
at height z and radial distance same as our test point x coordinate.
My path of integration around the ring will be parameterized by angle θ,
starting at the x axis and going counter-clockwise as seen from above. The
potential V at an observation point (ρ,φ,z) is given by
q
q(cid:48) = charge density
2πR
(cid:112)
r = (x−Rcosθ)2 +(y −Rsinθ)2 +z2 distance from ring to observation point
(cid:112)
= x2 +y2 +z2 +R2 −2xcosθ−2ysinθ expanded
(cid:112)
= ρ2 +z2 +R2 −2ρRcosθ aligning θ and φ
(cid:73)
1 (cid:16) q (cid:17) 1
V = Rdθ
4π(cid:15) 2πR r
(cid:73)
q 1
= dθ
8π2(cid:15) r
(cid:73)
q dθ
=
(cid:112)
8π2(cid:15) ρ2 +z2 +R2 −2ρRcosθ
(cid:73)
q 1 dθ
= √
(cid:112)
8π2(cid:15) ρ2 +z2 +R2 1−b2cosθ
(cid:115)
2ρR
b =
ρ2 +z2 +R2
We look at this, and recognize our K(k) integral in disguise.
2
1√ (cid:73) dθ
K(k) = 1+b2 √
4 1−b2cosθ
(cid:73)
dθ 4
√ = √ K(k)
1−b2cosθ 1+b2
(cid:32)(cid:114) (cid:33)
4 2b2
= √ K
1+b2 1+b2
2ρR
b2 =
ρ2 +z2 +R2
2ρR+ρ2 +z2 +R2
1+b2 =
ρ2 +z2 +R2
(cid:73)
q 1 dθ
V = √
(cid:112)
8π2(cid:15) ρ2 +z2 +R2 1−b2cosθ
(cid:32)(cid:114) (cid:33)
q 1 4 2b2
= √ K
(cid:112)
8π2(cid:15) ρ2 +z2 +R2 1+b2 1+b2
(cid:32)(cid:114) (cid:33)
q 1 2b2
= K
(cid:112)
2π2(cid:15) 2ρr+ρ2 +z2 +R2 1+b2
We now substitute for b2 to get a nice, standalone expression.
(cid:32)(cid:115) (cid:33)
q 1 4ρR
V(ρ,φ,z) = K
(cid:112)
2π2(cid:15) 2ρR+ρ2 +z2 +R2 2ρR+ρ2 +z2 +R2
(cid:32)(cid:115) (cid:33)
q 1 4ρR
= K
(cid:112)
2π2(cid:15) (ρ+R)2 +z2 (ρ+R)2 +z2
q 1
V(ρ,φ,z) = K(k)
(cid:112)
2π2(cid:15) (ρ+R)2 +z2
(cid:115)
4ρR
k =
(ρ+R)2 +z2
The voltage along the X axis is plotted in Figure 1. The voltage along
the Z axis is plotted in Figure 2.
3
Figure 1: Voltage Along X Axis for a 1 µC, 1 m Radius Loop
4
Figure 2: Voltage Along Z Axis for a 1 µC, 1 m Radius Loop
5
Static Electric Field
We are working in cylindrical coordinates. The gradient components are
(cid:126) (cid:126)
E = −∇V
∂V 1∂V ∂V
(cid:126)
∇V = (cid:126)a +(cid:126)a +(cid:126)a
ρ φ z
∂ρ ρ ∂φ ∂z
Repeating our potentials, we see V is independent of φ.
(cid:115)
4Rρ
k =
z2 +(R+ρ)2
q 1
V(ρ,φ,z) = K(k)
(cid:112)
2π2(cid:15) (R+ρ)2 +z2
The electric field components are then
∂V
E = −
ρ
∂ρ
E = 0
φ
∂V
E = −
z
∂z
We will have a number of terms in the derivative expression. We should
list some of these before we get started.
(cid:20) (cid:21)
∂K(k) E(k) K(k) ∂k
= −
∂ρ k(1−k2) k ∂ρ
(cid:20) (cid:21)
∂K(k) E(k) K(k) ∂k
= −
∂z k(1−k2) k ∂z
6
(cid:115)
4Rρ
k =
z2 +(R+ρ)2
(cid:115) √
∂k 1 4R 1 4Rρ(2(R+ρ))
= −
∂ρ 2 ρ(z2 +(ρ+R)2) 2(cid:0)z2 +(R+ρ)2(cid:1)3/2
(cid:20) (cid:21)
k 1 2(R+ρ)
= −
2 ρ z2 +(R+ρ)2
k (cid:20)z2 +(R+ρ)2 −2(R+ρ)ρ(cid:21)
=
2 ρ(z2 +(R+ρ)2)
k (cid:20) z2 +R2 −ρ2 (cid:21)
=
2ρ z2 +(R+ρ)2
(cid:115)
4Rρ
k =
z2 +(R+ρ)2
√
(cid:18) (cid:19)
∂k 4Rρ 1
= − 2z
∂z (cid:0)z2 +(R+ρ)2(cid:1)3/2 2
∂k zk
= −
∂z z2 +(R+ρ)2
Cylindrical Radial Field Component
We start with E .
ρ
q 1
V(ρ,φ,z) = K(k)
(cid:112)
2π2(cid:15) (R+ρ)2 +z2
∂V
E = −
ρ
∂ρ
(cid:32) (cid:33)
q ∂ 1
= − K(k)
(cid:112)
2π2(cid:15)∂ρ (R+ρ)2 +z2
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(cid:34) (cid:32) (cid:33) (cid:35)
q ∂ 1 1 ∂K(k)
E = − K(k)+
ρ 2π2(cid:15) ∂ρ (cid:112)(R+ρ)2 +z2 (cid:112)(R+ρ)2 +z2 ∂ρ
(cid:34)(cid:32) (cid:33) (cid:35)
q −(R+ρ) 1 ∂K(k)
= − K(k)+
(cid:112)
2π2(cid:15) (cid:0)(R+ρ)2 +z2(cid:1)3/2 (R+ρ)2 +z2 ∂ρ
Typesetting requires I re-arrange the leading constants to allow line breaks
(cid:32) (cid:33)
2π2(cid:15) −(R+ρ)
− E = K(k)+
q ρ (cid:0)(R+ρ)2 +z2(cid:1)3/2
1 (cid:20) E(k) K(k)(cid:21) k (cid:20) z2 +R2 −ρ2 (cid:21)
−
(cid:112)
(R+ρ)2 +z2 k(1−k2) k 2ρ z2 +(R+ρ)2
(cid:32) (cid:33)
−(R+ρ)
= K(k)+
(cid:0)(R+ρ)2 +z2(cid:1)3/2
1 (cid:20) E(k) (cid:21) 1 (cid:20) z2 +R2 −ρ2 (cid:21)
−K(k)
(cid:112)
(R+ρ)2 +z2 (1−k2) 2ρ z2 +(R+ρ)2
(cid:32) (cid:33)
−(R+ρ)
= K(k)−
(cid:0)(R+ρ)2 +z2(cid:1)3/2
(cid:32) (cid:33)
−(R+ρ) (cid:20) E(k) (cid:21) 1 (cid:20)z2 +R2 −ρ2(cid:21)
−K(k)
(cid:0)(R+ρ)2 +z2(cid:1)3/2 (1−k2) 2ρ (R+ρ)
Re-arranging our new common factor for typesetting, we have
(cid:18)2π2(cid:15)(cid:19)(cid:32)(cid:0)(R+ρ)2 +z2(cid:1)3/2(cid:33) (cid:20) E(k) (cid:21) 1 (cid:20)z2 +R2 −ρ2(cid:21)
−E = −K(k)+ −K(k)
ρ q (R+ρ) (1−k2) 2ρ (R+ρ)
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We are going to work on the right hand side for a little bit.
(cid:20) E(k) (cid:21) 1 (cid:20)z2 +R2 −ρ2(cid:21)
RHS = −K(k)+ −K(k)
(1−k2) 2ρ (R+ρ)
4ρR
k2 =
(R+ρ)2 +z2
(R−ρ)2 +z2
1−k2 =
(R+ρ)2 +z2
1 (R+ρ)2 +z2
=
1−k2 (R−ρ)2 +z2
(cid:20) (R+ρ)2 +z2 (cid:21) 1 (cid:20)z2 +R2 −ρ2(cid:21)
RHS = −K(k)+ E(k) −K(k)
(R−ρ)2 +z2 2ρ (R+ρ)
(cid:20) (R+ρ)2 +z2 (cid:21) 1 (cid:20)z2 +R2 −ρ2(cid:21)
RHS = −K(k)+ E(k) −K(k)
(R−ρ)2 +z2 2ρ (R+ρ)
(cid:20) (R+ρ)2 +z2 (cid:21)(cid:20)z2 +R2 −ρ2(cid:21)
= −K(k)+ E(k) −K(k)
(R−ρ)2 +z2 (2ρR+2ρ2)
(cid:20) z2 +R2 −ρ2(cid:21) (cid:20) (R+ρ)2 +z2(cid:21)(cid:20)z2 +R2 −ρ2(cid:21)
= −K(k) 1+ + E(k)
(2ρR+2ρ2) (R−ρ)2 +z2 (2ρR+2ρ2)
(cid:20)(R+ρ)2 +z2(cid:21) (cid:20) (R+ρ)2 +z2(cid:21)(cid:20)z2 +R2 −ρ2(cid:21)
= −K(k) + E(k)
(2ρR+2ρ2) (R−ρ)2 +z2 (2ρR+2ρ2)
(cid:20)(R+ρ)2 +z2(cid:21)(cid:18) (cid:20) z2 +R2 −ρ2 (cid:21)(cid:19)
RHS = −K(k)+ E(k)
2ρ(R+ρ) (R−ρ)2 +z2
(cid:18)2π2(cid:15)(cid:19)(cid:32)(cid:0)(R+ρ)2 +z2(cid:1)3/2(cid:33)
LHS = −E
ρ
q (R+ρ)
Re-uniting, and clearing common factors, we have
q 1 1 (cid:18) (cid:20) z2 +R2 −ρ2 (cid:21)(cid:19)
E = K(k)− E(k)
ρ 2πρ2π(cid:15)(cid:112)(R+ρ)2 +z2 (R−ρ)2 +z2
q (cid:18)K(k) (cid:20)E(k) z2 +R2 −ρ2 (cid:21)(cid:19)
= −
(cid:112)
4π(cid:15)ρ (R+ρ)2 +z2 π π (R−ρ)2 +z2
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Axial Field Component
∂V
E = −
z
∂z
(cid:32) (cid:33)
(cid:16) q (cid:17) ∂ 1
= − K(k)
(cid:112)
2π2(cid:15) ∂z (R+ρ)2 +z2
(cid:32) (cid:33)
(cid:18)2π2(cid:15)(cid:19) ∂ 1 1 ∂K(k)
− E = K(k)+
z (cid:112) (cid:112)
q ∂z (R+ρ)2 +z2 (R+ρ)2 +z2 ∂z
As before, we will work on the right hand side.
(cid:32) (cid:33)
(cid:18)2π2(cid:15)(cid:19) ∂ 1 1 ∂K(k)
− E = K(k)+
z (cid:112) (cid:112)
q ∂z (R+ρ)2 +z2 (R+ρ)2 +z2 ∂z
(cid:32) (cid:33)
−z 1 ∂K(k)
= K(k)+
(cid:112)
(cid:0)(R+ρ)2 +z2(cid:1)3/2 (R+ρ)2 +z2 ∂z
(cid:32) (cid:33)
−z
= K(k)
(cid:0)(R+ρ)2 +z2(cid:1)3/2
(cid:20) (cid:21)
1 E(k) K(k) ∂k
+ −
(cid:112)
(R+ρ)2 +z2 k(1−k2) k ∂z
(cid:32) (cid:33)
−z
= K(k)
(cid:0)(R+ρ)2 +z2(cid:1)3/2
(cid:20) (cid:21)(cid:18) (cid:19)
1 E(k) K(k) zk
+ − −
(cid:112)
(R+ρ)2 +z2 k(1−k2) k z2 +(R+ρ)2
(cid:32) (cid:33)
−z
= K(k)
(cid:0)(R+ρ)2 +z2(cid:1)3/2
(cid:32) (cid:33)
(cid:20) (cid:21)
z E(k)
− −K(k)
(cid:0)(R+ρ)2 +z2(cid:1)3/2 (1−k2)
(cid:32) (cid:33)
(cid:20) (cid:21)
z E(k)
= −
(cid:0)(R+ρ)2 +z2(cid:1)3/2 (1−k2)
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