Theoretische Physik 2: Elektrodynamik Home assignment 4 .Home assignment 4 Problem4.1...

download Theoretische Physik 2: Elektrodynamik Home assignment 4 .Home assignment 4 Problem4.1 Green’s reciprocation

of 12

  • date post

    26-Jun-2018
  • Category

    Documents

  • view

    219
  • download

    0

Embed Size (px)

Transcript of Theoretische Physik 2: Elektrodynamik Home assignment 4 .Home assignment 4 Problem4.1...

  • WiSe 2012 8.10.2012

    Prof. Dr. A.-S. SmithDipl.-Phys. Ellen FischermeierDipl.-Phys. Matthias Sabaam Lehrstuhl fr Theoretische Physik IDepartment fr PhysikFriedrich-Alexander-UniversittErlangen-Nrnberg

    Theoretische Physik 2: Elektrodynamik(Prof. A.-S. Smith)

    Home assignment 4

    Problem4.1 Greens reciprocation theorem

    Greens reciprocation theorem states that if is the potential due to a volume-charge density withina volume V and a surface-charge density on the conducting surface S bounding the volume V , and is the potential due to another charge distribution and , then

    V dV +

    S dS =

    V dV +

    S dS .

    a) Two infinite, grounded, parallel conducting planes are separated by a distance d. A point chargeq is placed between the planes. Use the reciprocation theorem of Green to prove that the totalinduced charge on one of the planes is equal to (q) times the fractional perpendicular distanceof the point charge from the other plane. Hint: As your comparison electrostatic problem withthe same surfaces choose one whose charge densities and potential are known and simple.

    b) Consider a potential problem in the half-space defined by z 0, with Dirichlet boundaryconditions on the plane z = 0 (and at infinity).(i) Write down the appropriate Greens function G(~x, ~x).(ii) Suppose that the potential on the plane z = 0 is specified to be = V inside a circle

    of radius a centered at the origin, and = 0 outside that circle. (1) Find an integralexpression for the potential at the point P specified in terms of the cylindrical coordinates(, , z). (2) Show that, along the axis of the circle ( = 0), the potential is given by

    = V

    (1 z

    a2 + z2

    )(3) Show that at large distances (2 + z2 a2) the potential can be expanded in a powerseries in (2 + z2)1, and that the leading terms are

    =V a2

    2

    z

    (2 + z2)3/2

    [1 3a

    2

    4(2 + z2)+

    5(32a2 + a4)

    8(2 + z2)2+ ...

    ].

    Verify that the results of parts (2) and (4) are consistent with each other in their commonrange of validity.

  • Problem4.2 Method of images

    a) Using the method of images, discuss the problem of a point charge q inside a hollow, grounded,conducting sphere of inner radius a. Find(i) the potential inside the sphere;(ii) the induced surface-charge density;(iii) the magnitude and direction of the force acting on q. Is there any change in the solution

    if the sphere is kept at a fixed potential V ? If the sphere has a total charge Q on its innerand outer surfaces?

    b) A conducting sphere is placed in a uniform external field. Calculate the potential and thesurface charge density .

    Problem4.3 Hollow cube

    A hollow cube has conducting walls defined by six planes x = 0, y = 0, z = 0, and x = a, y = a,z = a. The walls z = 0 and z = a are held at a constant potential V . The other four sides are at zeropotential.

    a) Find the potential (x, y, z) at any point inside the cube.b) Evaluate the potential at the center of the cube numerically, accurate to three significant figures.

    How many terms in the series is it necessary to keep in order to attain this accuracy?c) Find the surface-charge density on the surface z = a.

    Due date: Tuesday, 15.11.12

  • WiSe 2012 8.10.2012

    Prof. Dr. A.-S. SmithDipl.-Phys. Ellen FischermeierDipl.-Phys. Matthias Sabaam Lehrstuhl fr Theoretische Physik IDepartment fr PhysikFriedrich-Alexander-UniversittErlangen-Nrnberg

    Theoretische Physik 2: Elektrodynamik(Prof. A.-S. Smith)

    Solutions to Home assignment 4

    Solution of Problem4.1 Greens reciprocation theorem

    a)

    Fig. 1: A charge q between conducting planes.

    Let us place the two conducting planes perpendicu-larly to the z-axis, one through the origin, and theother through the point z = d, while the charge q is atthe distance a from the left plane (see Figure 1). Theplanes are grounded

    z=0

    = 0

    z=d

    = 0 ,

    while the point charge density is

    (~x) = q(x)(y)(z a) .

    On the planes, the surface charge densities 1, 2 arebeing induced.Consider the following problem: the charges q and q are uniformly distributed on the two infinitelylong, conducting planes. This configuration describes a parallel plate capacitor, for which we know thepotential at a general point to be (Figure 1)

    = Vd

    (z d) (whereV is the potential difference between the plates)

    z=0

    = V

    z=d

    = 0 ,

    We now use Greens reciprocation theorem. Here, the unprimed variables represent the quantities inthe case of there being a charge between the two planes, and the primed variables represent those inthe case of the parallel plate capacitor configuration, without a charge present in between. We get

    VdV +

    SdS =

    VdV +

    SdS

  • Now we calculate the different terms in the above equation.VdV =

    Vq(x)(y)(z a)V

    d(z d) dV = V q

    d(a d)

    SdS =

    z=0

    1dS +

    z=d

    2dS

    =0

    = V

    z=0

    1dS = V Q

    where Q is the total charge induced on the z = 0 plane, which we want to find.VdV = 0 (as = 0)

    SdS = 0 (as the planes are grounded).

    Putting the result for these four terms back into the Greens reciprocity relation, we get

    Q = qd

    (a d) ,

    where Q is the charge on the plane z = 0.

    b) (i)

    Fig. 2: Unit charge near a conducting, groundedplane.

    The Greens function for the specified region is equalto the solution for the potential in the problem of theunit charge near an infinite grounded conducting plane.If the unit charge is at the point with coordinates(x, y, z), and its image is at the point (x, y,z),then the Greens function reads

    G(~x, ~x) =1

    (x x)2 + (y y)2 + (z z)2

    1(x x)2 + (y y)2 + (z + z)2

    . (1)

    (ii) (1)

    Rewrite the Greens function in cylindrical coordinates

    x = cos

    y = sin

    z = z

    (x x)2 + (y y)2 + (z z)2 = ( cos cos)2 + ( sin sin)2 + (z z)2

    = 2 + 2 2 cos( ) + (z z)2

    G(~x, ~x) =1

    2 + 2 2 cos( ) + (z z)2 1

    2 + 2 2 cos( ) + (z + z)2.

    The general solution of the Dirichlet problem without the space distribution of a charge (x) reads

    (~x) = 14

    S

    (~x)G(~x, ~x)

    ndS .

    2

  • The normal is directed outside of the volume z 0 and therefore ~n = ~ez (see fig. 3). The derivationover the normal at the plane z = 0 is

    G(~x, ~x)

    n

    z=0

    = ~n G(~x, ~x)z=0

    = G(~x, ~x)

    z

    z=0

    =

    {(12) 2 (z z

    )(1)[2 + 2 2 cos( ) + (z z)2]3/2

    (12) 2 (z + z

    )

    [2 + 2 2 cos( ) + (z + z)2]3/2

    }z=0

    = 2z[2 + 2 2 cos( ) + z2]3/2

    At the surface at infinity the potential is zero. The final solution is then

    (, , z) =V z

    2

    20

    d a0

    d

    [2 + 2 2 cos( ) + z2]3/2. (2)

    Note that the potential distribution is azimuthally symmetric on the edge and hence the potential isindependent of the coordinate . So, in the above solution we can put = 0.(2)For = 0, the solution in (2) becomes

    (z) =V z

    2

    20

    d a0

    d

    [2 + z2]3/2= V

    (1 z

    a2 + z2

    ).

    (3)

    Fig. 3: Area of the disc shape at the potential Von the conducting, grounded plane.

    We will expand the solution (2) into its Taylor seriesin powers of 2 + z2. For 2 + z2 the expansion willconverge fast.

    [2 + 2 2 cos + z2]3/2

    =

    [2 + z2]3/21[

    1 + 22 cos

    2+z2

    ]3/2=

    [2 + z2]3/2

    {1 3

    2

    2 2 cos

    2 + z2

    +3 52 4

    (2 2 cos

    2 + z2

    )2+ ...

    }.

    Integration of the first term gives 20

    d a0

    d

    [2 + z2]3/2=

    a2

    [2 + z2]3/2(3)

    Integration of the second term gives

    32(2 + z2)[2 + z2]3/2

    20

    d a0(2 2 cos)d = 3a

    4

    4(2 + z2)[2 + z2]3/2(4)

    Integration of the third term gives

    15

    8(2 + z2)2[2 + z2]3/2

    20

    d a0(2 2 cos)2d

    =15

    8(2 + z2)2[2 + z2]3/2

    (3a6 + a42

    )(5)

    3

  • utting in the results from (3), (4) and (5) into (2), we get

    (, z) =V a2

    2

    z

    (2 + z2)3/2

    [1 3a

    2

    4(2 + z2)+

    5(32a2 + a4)

    8(2 + z2)2+ ...

    ]. (6)

    Verification. Does the solution (6) for = 0 match with the one we got in part (2) for z a? From(6),

    ( = 0, z) =V a2

    2

    1

    z2

    [1 3

    4

    (az

    )2+

    5

    8

    (az

    )4+ ...

    ].

    The solution from part (2) for z a becomesz

    a2 + z2=

    11 +

    (az

    )2 = 1 12 (az)2 + 38 (az)4 + ...(z) = V

    [1

    (1 1

    2

    (az

    )2+

    3

    8

    (az

    )4 5

    16

    (az

    )6+ ...

    )]=V a2

    2

    1

    z2

    [1 3

    4

    (az

    )2+

    5

    8

    (az

    )4+ ...

    ].

    Hence we have shown that the solutions match in the specified range.

    Solution of Problem4.2 Method of images

    a) (i) The solution for the potential inside a spherical cavity is of the form (Fig. 4.)

    (~x) =q

    r2 + d2 2rd cos +

    qr2 + d2 2rd cos

    ,

    where q is the value and d is the position of the image charge. Both q and d have to beevaluated from the boundary condition

    (~x)|r=a = 0 .

    Using the assumed form of the potential with the boundary condition,

    (~x)|r=a =q

    a2 + d2 2ad cos +

    qa2 + d2 2ad cos

    = 0

    = qa2 + d2 2ad cos = q

    a2 + d2 2ad cos

    =[q2(a2 + d2) q2(a2 + d2)

    ]+ 2a cos

    [dq2 dq2

    ]= 0

    z d

    0d

    n q q

    Fig. 4: Point charge inside of the sphericalcavity in the conductor.

    Functions 1 and cos are linearly independent,thus