topic 07 electrostaticfieldproblems

8/6/2019 Topic 07 ElectrostaticFieldProblems

1/23

Electrostatic Field Problems:

Spherical Symmetry

EE 141 Lecture Notes

Topic 7

Professor K. E. OughstunSchool of Engineering

College of Engineering & Mathematical Sciences

University of Vermont

2009


2/23

Motivation


3/23

Spherical Coordinates

Coordinate transformations

x= rsin cos , y = rsin sin , z= rcos (1)

r =x2 +y2 + z2, = arctan

x2 +y2

z

, = arctan

yx

(2)

with 0 r , 0 , and 0 < 2.


4/23

Unit Basis Vectors in Spherical Coordinates

Unit basis vectors

1r = 1r(, ) = 1x sin cos + 1y sin sin + 1z cos ,1 = 1(, ) = 1x cos cos + 1y cos sin 1z sin , (3)1 = 1(, ) = 1x sin + 1y cos ,

where (Orthogonality Relations)

1r 1 = 1,1 1 = 1r, (4)1 1r = 1.

Problem 13. Using the relations in Eq. (3), determine expressions forthe unit basis vectors 1x, 1y, and 1z in terms of the unit basis

vectors 1r, 1, and 1.

Problem 14. Determine the direct coordinate transformation inmatrix notation from c lindrical to s herical coordinates.


5/23

Vectors in Spherical Coordinates

Any vector V may be expressed in spherical polar coordinates as

V =1VV =

1rVr +

1V +

1V, (5)

where

Vr = 1r V= (1x sin cos + 1y sin sin + 1z cos )

(1xVx + 1yVy + 1zVz)

= Vx sin cos + Vy sin sin + Vz cos ,

V = 1 V= (1x cos cos + 1y cos sin + 1z sin )(1xVx + 1yVy + 1zVz)

= Vx cos cos + Vy cos sin + Vz sin , (6)V = 1 V = (1x sin + 1y cos ) (1xVx + 1yVy + 1zVz)

= Vx sin + Vy cos ,with magnitude

V = V = V V = V2 + V2 + V2. 7


6/23

Position Vector in Spherical Coordinates

Position Vector of a point P = P(r, , ) is given by

R = OP= 1r(, )r, (8)

where r =x2 +y2 + z2.

Notice that this position vector does not have either a 1 or 1component. Nevertheless, the orientation of the radial unit vector1r = 1r(, ) depends upon the and coordinates of the point P.


7/23

Spherical Coordinates - Differential Elements

Differential elements of length along the 1r, 1, & 1-directions:

dr = dr, d = rd, d = rsin d.


8/23

Differential Length, Surface Area, & Volume

The vector differential element of length:

d = 1rdr+ 1rd + 1rsin d. (9)

Fundamental quadratic form or metric form

d2 = dr2 + r2d2 + r2 sin2 d2. (10)

Differential elements of surface area:

-Spherical Surface: dsr = 1r (dd) = 1rr2 sin dd.

r-Conical Surface: ds = 1 (drd) = 1rsin drd.

r-Planar Surface: ds = 1 (drd) = 1rdrd.

Differential element of volume:

dV = drdd = r2 sin drdd. (11)


9/23

Distance Between Two Points

Coordinates ofP1(x1,y1, z1) = P1(r1, 1, 1):

x1 = r1 sin 1 cos 1, y1 = r1 sin 1 sin 1, z1 = r1 cos 1.

Coordinates ofP2(x2,y2, z2) = P2(r2, 2, 2):

x2 = r2 sin 2 cos 2, y2 = r2 sin 2 sin 2, z2 = r1 sin 1 sin 1.

Distance between P1 & P2 is then given by Pythagoreans theorem as

d =

(r2 sin 2 cos 2 r1 sin 1 cos 1)2

+(r2 sin 2 sin 2

r1 sin 1 sin 1)

2

+(r1 sin 1 sin 1 r1 cos 1)21/2

=

r21 + r

22 2r1r2

cos 1 cos 2 + sin 1 sin 2 cos(2 1)

1/2

.

(12)

V t Diff ti l O t i S h i l P l


10/23

Vector Differential Operators in Spherical Polar

Coordinates

Gradient Operator

= 1r r

+ 11

r

+ 1

1

rsin

(13)

The gradient of a scalar function of position f(r) = f(r, , ) is thengiven by

f= 1rfr

+ 11

r

f

+ 1

1

rsin

f

. (14)

Laplacian Operator

2 = 1r2

r

r2

r

+

1

r2 sin

sin

+

1

r2 sin2

2

2(15)

V t Diff ti l O t i S h i l P l


11/23

Vector Differential Operators in Spherical Polar

Coordinates

Vector function of position F(r) = F(r, , ) = 1rFr + 1F + 1Fhas divergence

F = 1r2

r(r2Fr) +

1

rsin

(F sin ) +

1

rsin

F

(16)

and curl

F = 1r 1rsin

(F sin ) F

+1

1rsin

Fr

sin r

(rF)

(17)

+11

r

r(rF) Fr

.

Fi ld f P i Ch (S b P i )


12/23

Field of a Point Charge (Symmetry about a Point)

Consider a point charge Q located at the origin of coordinates (ifnot, the origin of coordinates can always be shifted to that point).

The spherical symmetry of the problem then requires that

E(r) = 1rE(r).



13/23


Application of Gauss law to a concentric spherical surfaceS

withradius r > 0 surrounding the point charge Q at the origin O yields

Q

0=

S

E(r)1r 1rda = E(r)S

da = 4r2E(r),

so that

E(r) = 1rE(r) = 1rQ

40r2, r > 0

The absolute potential due to the point charge is then given by

V(r) =

r

1rE(r) 1rdr = Q40r

, r > 0.



14/23


0 1 2 3 4

r

0

Q/40

E ~ 1/r2

V ~ 1/r

Fi ld f U if S h i l Ch Di t ib ti


15/23

Field of a Uniform Spherical Charge Distribution

Consider determining the field due to a uniform spherical chargedistribution of radius r0 > 0 with volume charge density (r) = 0 for

r r0 centered on the origin, where (r) = 0 for r > r0.

The spherical symmetry of the problem then requires that

E(r) =

1rE(r).

Application of Gauss law to a concentric sphere of radius r centeredat the origin then yields

4r2E(r) = 10

V

(r)d3r = 10 43 r3 0, r r04

3r30 0, r > r0

Notice that if 0 is positive, then E is directed radially outward fromthe origin, whereas if 0 is negative, then E is directed radially inwardtoward the origin.

Fi ld f U if S h i l Ch Di t ib ti


16/23


The total charge Q contained in the spherical charge distribution is

Q=4

3 r30 0,

so that

0 =3Q

4r30.

The radial component of the electric field vector E(r) = 1rE(r) isthen given by

E(r) = Q

40r30r, r r0

Q4

0r2

, r > r0

Notice that measurements of the electric field due to a sphericallysymmetric charge distribution localized in space (r r0) for r > r0are independent of the radius of the charge distribution.In the limit as r

0 0, the field due to a point charge Q at the origin

is obtained.

Field of U ifo S he ic l Ch ge Dist ib tio


17/23


The absolute potential [Eq. (4.8)] due to the uniform spherical

charge distribution is then given by (with d = 1rdr)

V(r) =

r

E(r)dr.

For r r0,V(r) =

Q

40r

dr

r2 =Q

40r

which is the same as that due to a point charge Q at the origin.For r r0,

V(r) = Q40

1r30

r0r

rdr+ r0

drr2

=Q

80r30 r20 r2

increase in potential above the surface value+

Q

40r0 potential at the sphere surface.



18/23


0 r0

2r0

r

0

Q/40r

0

Q/40r

0

3Q/80r0

2

E(r)

V(r)



19/23


The potential V(r) may also be determined using Poissons &

Laplaces equations directly, taking advantage of the sphericalsymmetry of the source charge distribution.

Inside the uniform charge distribution (r r0), Poissons equation

2V(r) =

0/0 becomes

1

r2

r

r2

V

r

= 0

0=

r

r2

V

r

= 0

0r2

r2V

r=

0

30r3 + C =

V

r=

0

30r+

C

r2

V(r) = 060

r2 Cr

C = 0+D = V(r) = Q

80r30r2 + D.



20/23


Outside the uniform charge distribution (r r0), Laplaces equation2V(r) = 0 becomes1

r2

rr2 V

r = 0 =

rr2 V

r = 0 r2

V

r= A = V

r=

A

r2

V(r) = Ar

+ B; V() 0 = B= 0

V(r) = Ar

.



21/23


Continuity ofV(r) at r = r0 requires that

Ar0

= Q80r0

+ D,

while continuity of V/r [i.e., continuity ofE(r)] at r = r0 requires

A

r20= Q

80r20= A = Q

80

so thatD=

Q

80r0 A

r0=

Q

80r0+

Q

40r0.



22/23


The electrostatic potential is then given by

V(r) =Q

80r30

r20 r2

+

Q

40r0; r r0

V(r) =Q

40r

; r

r0

so that the electric field intensity E(r) = V(r) = 1rV/r is

E(r) = 1rQ

40r3

0

r ; r

r0

E(r) = 1rQ

40r2; r r0

in agreement with the result obtained using Gauss law.

Take Home Exam Problem 2


23/23

Take Home Exam Problem 2

A spherical region of radius a > 0 situated in free space contains avolume charge density given by

(r) = 0

1 + r2

; r a,with (r) = 0 for r > a, where 0 and are constants.

1 (20 points) Utilize Gauss law together with the inherent

symmetry of the problem to derive the electrostatic field vectorE(r) both inside and outside the spherical charge region.

2 (40 points) Use both Poissons and Laplaces equations todirectly determine the electrostatic potential V(r) both inside

and outside the spherical region. From this potential function,determine the electrostatic field vector E(r).3 (40 points) Determine the value of the parameter for which

the electrostatic field vanishes everywhere in the region outsidethe spherical charge region (r > a). Plot Er(r) and V(r) as a

function of r for this value of .

topic 07 electrostaticfieldproblems

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