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Ηλεκτρικό ∆ίπολο - Εύρεση Ηλεκτρικού Πεδίου

(ανεξάρτητα του συστήµατος συντεταγµένων)

Χρησιµοποιώντας τις προσεγγίσεις του ηλεκτρικού διπόλου, το ηλεκτροστατικό δυναµικό στην τυχαία ϑέση

~r (ϑεωρώντας το ηλεκτρικό δίπολο στο κέντρο του συστήµατος συντεταγµένων και κατά µήκος του άξονα z),

δίδεται από την σχέση

Φ(~r) =~p · ~r

4π�r3,

όπου ~p = q~d = qdı̂z η ϱοπή του ηλεκτρικού διπόλου. Το ηλεκτρικό πεδίο, ~E, δίδεται από την κλίση του

δυναµικού και εποµένως δίδεται από

~E = −~∇Φ = 14π�

~∇(~p · ~rr3

).

Για το υπολογισµό του ηλεκτρικού πεδίου απαιτούνται τα gradient:

~∇(

1

r3

)=

∂

∂x

(1

r3

)ı̂x +

∂

∂y

(1

r3

)ı̂y +

∂

∂z

(1

r3

)ı̂z =

−3~rr5

,

~∇ (~p · ~r) = (~p · ~∇)~r + (~r · ~∇)~p+ ~p× (~∇× ~r) + ~r × (~∇× ~p) =

= (~p · ~∇)~r = ~p,

εφόσον r = (x2 + y2 + z2)1/2, (~r · ~∇)~p = 0, ~∇ × ~r = 0, και ~∇ × ~p = 0. Χρησιµοποιώντας τα παραπάνωgradient το ηλεκτρικό πεδίο δίδεται από την κάτωθι εξίσωση

~E(~r) =3(~p · ı̂r )̂ır − ~p

4π�r3.

Στην περίπτωση που το ηλεκτρικό δίπολο ϐρίσκεται σε τυχαία ϑέση αντικαθιστούµε το διάνυσµα ϑέσης ~r

µε το ~R = ~r − ~r ′ όπου το ~r ′ αντιστοιχεί στην ϑέση του ηλεκτρικού διπόλου. Εποµένως το δυναµικό και τοηλεκτρικό πεδίο δίδονται από τις εξισώσεις :

Φ(~r) =~p · ~R

4π�R3=

~p · ı̂R4π�R2

,

~E(~r) =3(~p · ı̂R)̂ıR − ~p

4π�R3,

όπου ı̂R = ~R/R.

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NATIONAL TECHNICAL UNIVERSITY OF ATHENS

School of Electrical & Computer EngineeringDivision of Electromagnetic Appl., Electro-Optics, & Electronic Materials

9 Iroon Polytechniou St., Zografou, Athens 15780Tel. (30) 210-7722479, Fax. (30) 210-7722281

E-mail: [email protected]

Spherical Cavity in a Spherical Conductor (October 20, 2019)†

Problem Statement

Assume a spherical conductor of radius a in vacuum. Inside the conductor there is a spherical cavity

(an air bubble) of radius b. The center of the cavity and its radius are such that the whole cavity

is inside the interior of the conductor as shown in Fig. 1. Assume that inside the spherical cavity is

a point charge q positioned at the center of the cavity (as shown). Assume that the permittivity is

everywhere equal to the permittivity of freespace �0.

(a) Assume that the conductor is uncharged before the positioning of the point charge q inside the

cavity. Find the electrostatic potential and the electric field everywhere, i.e., inside the cavity, in the

interior, and in the exterior of the spherical conductor.

(b) Repeat (a) in the case that the spherical conductor is always retained at a potential U0 (for example

when it is connected to a battery of potential U0).

�

�

�

�

�

�

Figure 1: A spherical conductor of radius a has in its interior a cavity of radius b with the position of thecavity and its radius such that the whole cavity is inside the volume of the spherical conductor (as shown).Assume a point charge q at the center of the cavity.

† c©2019 Prof. Elias N. Glytsis, Last Update: October 20, 2019

Solution

(a) It is easier to start for the interior of the spherical conductor. As it is known a conductor in a

static situation does not have in its interior electric field and its volume (and surface) is equipotential.

Therefore, in the interior of the conductor (excluding the volume of the cavity) the potential is constant.

Assume this potential Φ0 (Φ0 has not been determined yet). Since the interior of the conductor is

equipotential the electric field should be zero (excluding the cavity area). In a second step let’s examine

the cavity. Inside the cavity is a point charge at the center of it. The spherical boundary of the cavity

is equipotential and should have potential Φ0 since it is part of the interior of the spherical conductor.

Therefore, the solution for the potential inside the cavity is

Φ(rb) − Φ(rb = b) =q

4π�0

(

1

rb−

1

b

)

, for 0 < rb ≤ b,

Φ(rb) =q

4π�0

(

1

rb−

1

b

)

+ Φ0, for 0 < rb ≤ b. (1)

In the above equation rb is the radial distance from the center of the cavity. This of course implies an

electric field inside the cavity that can be easily determined from the gradient of the electric potential,

i.e., ~E = −~∇Φ = (q/4π�0)(1/r2

b)̂ırb. This field will induce a surface charge on the boundary of the

cavity which the inner boundary surface of the spherical conductor. Due to the spherical symmetry

of the cavity the surface charge density, σb, should be uniform. This can be easily determined from

the boundary condition on the cavity boundary: σb = ı̂rb • (0−~D|b) = −(q/4πb

2). However, since the

spherical conductor is initially uncharged, a surface charge density σa will be induced at the outer

surface of the spherical conductor, such as the total charge should be zero. Since the surface charge

density, due to spherical symmetry, should be uniform, then

σa(4πa2) + σb(4πb

2) = 0 =⇒ σa = +q

4πa2. (2)

At this point let’s examine the exterior region of the spherical conductor. The potential due to the

spherical symmetry should depend only in the radial distance ra from the center of the conductor (for

ra ≥ a). The corresponding Laplace’s equation takes the following form:

1

r2a

d

dra

(

r2adΦ

dra

)

= 0 =⇒ Φ(ra) = −C1ra

+ C2, (Ra ≥ a),

but Φ(ra → ∞) → 0 =⇒ C2 = 0,

Φ(ra) = −C1ra

, (ra ≥ a).

In the above expression ra is the radial distance from the center of the spherical conductor. Then the

electric field outside the conductor and the induced surface charge density are related as follows:

σa = ı̂ra • (~D|a − 0) = −�a

C1a2

=⇒ C1 = −σaa

2

�0= −

q

4π=⇒

Φ(ra) =q

4π�0ra, (ra ≥ a) and Φ0 = Φ(ra = a) =

q

4π�0a.

Then all the parameters have been defined and the electric potential is summarized as follows:

Φ =

q

4π�0

(

1

rb−

1

b

)

+q

4π�0a, for rb ≤ b,

q

4π�0afor ra ∈ interior of conductor,

q

4π�0ra, for ra ≥ a.

(3)

The corresponding electric field is given by

~E =

q

4π�0r2

b

ı̂rb, for rb < b,

0 for ra ∈ interior of conductor,

q

4π�0r2aı̂ra, for ra > a.

(4)

It is reminded that at rb = b and ra = a the electric field is discontinuous due to the surface charge

densities σb and σa, respectively.

(b) In the second case the spherical conductor is always retained in a constant potential U0. In this

case electric charge can either come from or can drop into the battery source. The solution to the

interior of the conductor (excluding the cavity) is the same as previously with the only difference that

Φ0 = U0 while the interior electric field is zero. Inside the cavity the solution is also identical to the

one shown in Eq. (1) with Φ0 = U0. Therefore, the only solution that changes is the region outer of

the spherical conductor. The solution for the electric potential for ra >≥ a is again Φ(ra) = −C1/ra.

However, now continuity of the potential at ra = a gives C1 = −U0a. Then, the solution for the

electric potential is summarized as follows:

Φ =

q

4π�0

(

1

rb−

1

b

)

+ U0, for rb ≤ b,

U0 for ra ∈ interior of conductor,

U0a

ra, for ra ≥ a.

(5)

The corresponding electric field is the given by:

~E =

q

4π�0r2

b

ı̂rb, for rb < b,

0 for ra ∈ interior of conductor,

U0a

r2aı̂ra, for ra > a.

(6)

It is worth mentioning that again that the surface charge density σb = −q/4πb2 as in case (a). However,

the surface charge density σa = ı̂ra • (~D|a − 0) = �0U0/a. The last surface charge density remained the

same even before the point charge q was positioned at the center of the spherical cavity. Therefore,

the total charge of the conductor before and after the insertion of q is not the same. The difference

was accommodated by the battery source that was connected to the spherical conductor (actually the

induced charge σb(4πb2) = −q came from the battery source since σa must not change.

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ΠΑΡΑ∆ΕΙΓΜΑΤΑ ΤΗΣ ΜΕΘΟ∆ΟΥ ΤΩΝ ΡΟΠΩΝ

ΣΤΗΝ ΗΛΕΚΤΡΟΣΤΑΤΙΚΗ †

1. Ευθύγραµµος Κυλινδρικός Αγωγός Μήκους 2L & Ακτίνας a

x direction (meters)-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5

Lin

e ch

arg

e (p

C/m

)

7.5

8

8.5

9

9.5

102L = 1m, ǫ r = 1, N = 15, a = 0.001 m, U = 1 V, Q = 8.5638 pC, Q1 = 8.4295 pC

Simple CoefficientsMore Accurate Coefficients


Lin

e ch

arg

e (p

C/m

)

7.5

8

8.5

9

9.5

10

10.5



†©2016 Prof. Elias N. Glytsis, Last Update: Oct. 18, 2016


Lin

e ch

arg

e (p

C/m

)

7.5

8

8.5

9

9.5

10

10.5

11

11.5




Lin

e ch

arg

e (p

C/m

)

7

8

9

10

11

12

13




Lin

e ch

arg

e (p

C/m

)

7

8

9

10

11

12

13

14

15




Lin

e ch

arg

e (p

C/m

)

-20

-10

0

10

20

30

40



Σχήµα 1: Γραµµική πυκνότητα ϕορτίου ευθύγραµµου κυλινδρικού αγωγού µήκους 1 µέτρου και ακτίνας 0.001 µέτρου. Ν =15, 25, 50, 100, 150, και 500. Παρατηρούµε ότι για Ν=500 η µέθοδος αποκλίνει γιατί παύουν να ισχύουν οι προϋποθέσεις τουγραµµικού, µονοδιάστατου, αγωγού.

2. Τετραγωνική Επίπεδη Αγώγιµη Πλάκα ∆ιαστάσεων 2L × 2L

0.5

2Lx = 1m, 2Ly = 1m, ǫ r = 1, Nx = 15, Ny = 15, U = 1 V, Q = 40.1823 pC

x direction (meters)

0

-0.5-0.5y direction (meters)

0

0

50

100

150

0.5

Su

rfac

e ch

arg

e (p

C/m

2)

0.5



0


0

0

50

100

150

200

0.5

Su

rfac

e ch

arg

e (p

C/m

2)

0.5



0


0

400

0

100

200

300

0.5

Su

rfac

e ch

arg

e (p

C/m

2)

0.5



0


0

0

100

200

300

400

500

0.5

Su

rfac

e ch

arg

e (p

C/m

2)

0.5



0


0

200

300

400

500

0

100

0.5

Su

rfac

e ch

arg

e (p

C/m

2)

Σχήµα 2: Τετραγωνική πλάκα διαστάσεων 1 µέτρου × 1 µέτρου. Nx = Ny = 15, 25, 50, 75, και 100.

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