Download Physics 417G : Solutions for Problem set 1

Physics 417G : Solutions for Problem set 1
Due : January 22, 2016
Let us refresh our memory!
Here are some problems with vector calculus and spherical objects in Electrostatics.
Please show all the details of your computations including intermediate steps.
1
Problem 1
~ × B)
~ × (C
~ × D)
~ = B(
~ A
~·C
~ × D)
~ − A(
~ B
~ ·C
~ × D)
~ .
a) Prove (A
b) Assuming ~v are a scalar and a vector as functions of three coordinates ~v = x̂vx (x, y, z)+ ŷvy (x, y, z)+
ẑvz (x, y, z), where x̂ is a unit vector along x−direction. Compute
~ · (∇
~ × ~v ) ,
i) ∇
~ × (∇
~ × ~v ) .
ii) ∇
~ ·∇
~ of the following function
Now, compute the Laplacian ∇
iii) ~va = x2 yx̂ + 3x2 y 2 ŷ − 2x2 z ẑ .
c) Here we practice derivatives
p for some functions with radial dependence. Let us consider the position
x2 + y 2 + z 2 , a scalar function f (r) with its derivative f 0 (r) = df /dr.
vector ~r = rr̂ with r =
Compute the following derivatives
~ ,
i) ∇r
~ × ~r .
ii) ∇
For a vector function ~g (r) (only a function of r), evaluate the following expressions in terms of
r̂, r, ~g , and ~g 0 , a derivative with respect to r,
~ · ~g ,
iii) ∇
~ × ~g ,
iv) ∇
~ · (~r × ~g ) ,
v) ∇
~ × (~r × ~g ) .
vi) ∇
Sol: a) We are going to use the index notation. The expression is a vector and we consider ith
component in terms of the index notation. Then
~ × B)
~ × (C
~ × D)]
~ i = ijk jlm kab Al Bm Ca Db = (δkl δim − δkm δil )kab Al Bm Ca Db
[(A
~ A
~·C
~ × D)
~ − A(
~ B
~ ·C
~ × D)]
~ i.
= Ak Bi kab Ca Db − Ai Bk kab Ca Db = [B(
~ × D)
~ k=
The first expression in the second line has Bi and Ai which are vectors. Ak kab Ca Db = Ak (C
~·C
~ × D,
~ where k is summed over as a dot product.
A
b)
~ · (∇
~ × ~v ) = ijk ∂i ∂j vk = 0 ,
i) ∇
~ × (∇
~ × ~v )]i = ijk klm ∂j ∂l vm = (δil δjm − δim δjl )∂j ∂l vm
ii) [∇
~ ∇
~ · ~v ) − ∇
~ 2~v ]i .
= ∂j ∂i vj − ∂j ∂j vi = [∇(
and
~ 2~va = (∂x2 + ∂y2 + ∂z2 )(3x2 y 2 x̂ − 2x2 z ŷ + x2 yẑ) = (6x2 + 6y 2 )x̂ − 4z ŷ + 2yẑ .
iii) ∇
c) Let us compute one by one.
~ i = ∂i (xj xj )1/2 = (∂i xk )xk (xj xj )−1/2 =
i) [∇r]
~ × ~r]i = ijk ∂j xk = ijk δjk = 0 .
ii) [∇
xi
,
r
~ = r̂ ,
∇r
The last equality is true because the indices j and k are antisymmetric in ijk , while symmetric in δjk
and xj xk . For vector ~g , we also evaluate one by one.
~ · ~g = ∂i gi = (∂i r)g 0 = xi gi0 = r̂ · ~g 0 ,
iii) ∇
i
r
~ × ~g ]i = ijk ∂j gk = ijk (∂j r)g 0 = ijk xj g 0 = [r̂ × ~g 0 ]i ,
iv) [∇
k
r k
~ × ~g = r̂ × ~g 0 ,
∇
xj xi gk0
~
v) ∇ · (~r × ~g ) = ijk ∂i (xj gk (r)) = ijk [xj ∂i gk (r) + gk ∂i xj ] = ijk
+ gk δij = 0 ,
r
~ × (~r × ~g )]i = ijk ∂j (klm xl gm ) = [δil δjm + δim δjl ][(∂j gm )xl + gm ∂j xl ]
vi) [∇
xj
xj
= (∂j gj )xi + gj ∂j xi − (∂j gi )xj − gi ∂j xj = xi gj0 + gi − ( gi0 )xj − gi δjj (= 3) ,
r
r
0
~
r
(~
r
·
~
g
)
0
~ × (~g × ~r) = −r~g − 2~g +
.
∇
r
where in j) we use the fact that is antisymmetric in all the indices while xk xi , δik are symmetric. If
you multiply symmetric and antisymmetric parts and summed over, the net result is 0. Throughout
x
the computations, we use ∂j gk (r) = rj gk0 (r).
2
Problem 2
a) Let us consider one representation of a delta
function in spherical coordinate, which is a variation
√
2
(3)
2
~
of ∇ (1/r) = −4πδ (~r). We replace r by r + 2 and try to see what happens as → 0. Consider
a function
1
1 ~2
√
D(r, ) = − ∇
.
2
4π
r + 2
i) Compute D(r, ) and integrate it over all space.
ii) Check that D(0, ) → ∞ as → 0 and D(r 6= 0, ) → 0 as → 0.
These properties demonstrate D(r, ) behaves δ (3) (~r) as → 0.
Note: for performing an integral, you can find it in the integral table or with some programs such as
Mathematica. Yet, please try to show it explicitly.
b) Find the divergence of the function
r̂
.
r
First compute it directly by using
the divergence
expression in spherical coordinate. Test your result
R
H
~ · ~v = ~v · dS,
~ where dS
~ is the surface element. Is there a delta
using the divergence theorem d3 x∇
function at the origin, as there was for ~r/r2 ?
c) What is the general formula for the divergence of rn r̂ for n > −2?
R
H
~ × ~v = − ~v × dS.
~
d) Find the curl of rn r̂. Test your conclusion using d3 x∇
~v =
Sol: a) i) We work on spherical coordinate, where the Laplacian of a function A(r) is given by
∇2 A(r) =
1
1
1
∂r r2 ∂r A(r) + 2
∂θ (sin θ∂θ A(r)) + 2 2 ∂φ2 (A(r)) ,
2
r
r sin θ
r sin θ
and thus we have
1
1
1
r
1
r3
2
2
D(r, ) = −
=
∂r r ∂r √
∂r r (−1) 2
∂r
=−
4πr2
4πr2
4πr2
(r + 2 )3/2
(r2 + 2 )3/2
r 2 + 2
2
3
=
,
2
4π(r + 2 )5/2
where we used that D is a function of r only.
R 2π
Rπ
Let us compute the integral. The angular part of the integral 0 dφ 0 sin θdθ = 4π. Thus
Z π/2
Z 1
Z ∞
Z ∞
r2 dr
2
2 1
4π
D(r, )r2 dr = 32
=
3
sin
θ
cos
θdθ
=
3
x2 dx = 1 .
2 0
(r2 + 2 )5/2
0
0
0
We change the coordinate as r = tan θ and also x = sin θ in the last two steps.
ii) It is straight forward to see
3
→ ∞, for → 0 ,
4π3
32
D(r, = 0) =
→ 0, for r 6= 0 .
4π(r2 + 2 )5/2
D(r = 0, ) =
b) For ~v , we have
~ · ~v = 1 ∂ r2 vr = 1 ∂
∇
r2 ∂r
r2 ∂r
r
21
r
=
1
.
r2
We can test the result with the divergence theorem for the sphere with radius R as
Z
Z
Z R
1
3 ~
d x∇ · ~v = sin θdθdφ
drr2 2 = 4πR ,
r
0
Z
Z
2
~ · ~v = R sin θdθdφr̂ · r̂ = 4πR .
dS
R
~ = r̂r2 sin θdθdφ. Thus there are no delta functions at the origin.
Where we used dS
c) The divergence of rn r̂ for n > −2 can be computed as
~ · (rn r̂) = 1 ∂ r2 rn = (n + 2)rn−1 .
∇
2
r ∂r
~ × ~v = ∇
~ × (rn r̂) = 0 because the curl of the radial component is related to
d) One can easily check ∇
H
H
~ = (rn r̂) × r̂R2 sin θdθdφ = 0 because r̂ × r̂ = 0.
angular derivatives. And also we can show ~v × dS
This should be the case geometrically because any radially diverging function does not have a circulating component. Thus there is not delta function here either!
3
Problem 3
Consider a spherical surface of radius R, which carries a uniform charge density σ, as in the figure.
a) What is the total charge Q?
b) Using Gauss law, find the electric field inside (r < R) and outside (r > R) of the spherical shell in
terms of the total charge Q.
c) By direct integral of the electric field expression,
Z
1
ρ(~x0 )
3 0
E=
r 2 r̂ d x .
4π0
Find the electric field a distance z from the center of a spherical surface of radius R for both z < R
(inside) as well as z > R (outside) in terms of the total charge Q.
Sol: a) The total charge is Q = 4πR2 σ.
H
~ · dS
~ = 4πr2 E(r) = Qenc (with spherical symmetry) tells us that
b) Gauss law in its integral form E
0
0
r<R
~ =
E
.
Q r̂
r>R
4π0 r 2
c) The direct integration gives the same answer. For the problem in hand, the electric field point in
z direction. We can solve the integral with the information r~ = ~x − ~x0 = z ẑ − r̂R, r 2 = R2 + z 2 −
2Rz cos θ, ρ(~x0 ) = δ(r − R)σ. Then we get
1
E=
4π0
Z
ρ(~x0 )
r
2
2π
σ
d x =
4π0
r̂
3 0
Z
2π
Z
dφ
0
π
sin θdθ
0
R2 (z ẑ − r̂R)
(R2 + z 2 − 2Rz cos θ)3/2
π
R2 (z ẑ − R(cos θẑ + sin θ cos φx̂ + sin θ sin φŷ))
dφ
sin θdθ
(R2 + z 2 − 2Rz cos θ)3/2
0
0
Z π
Z
2πσ 1
2πσ
R2 (z − R cos θ)ẑ
R2 (z − Ru)ẑ
=
=
sin θdθ 2
du 2
2
3/2
4π0 0
4π0 −1 (R + z 2 − 2Rzu)3/2
(R + z − 2Rz cos θ)
Z
2πσR2 ẑ 1
1
1
z 2 − R2
1
=
du
+
4π0
2z (R2 + z 2 − 2Rzu)3/2
2z (R2 + z 2 − 2Rzu)1/2
−1
1
2
2πσR2 ẑ 1
zu − R
z−R
−z − R
= 2πσR ẑ 1
=
−
4π0 z 2 (R2 + z 2 − 2Rzu)1/2 4π0 z 2 |z − R|
|z + R|
σ
=
4π0
Z
Z
−1
where we use
R 2π
0
dφ sin φ =
R 2π
0
dφ cos φ = 0 in the third line, u = cos θ in the fourth line. This gives
~ =
E
4
0
z<R
z>R
Q ẑ
4π0 z 2
.
Problem 4
A spherical shell of radius R, carrying charge Q, is surrounded by a thick concentric metal shell (inner
radius a, outer radius b, as in the figure), which carries a net charge q.
a) Find the surface charge densities at each surface, σR at R, σa at a, and σb at b.
b) Find the potential at the center, using infinity as the reference point.
Now the outer surface is touched to a grounding wire, which lowers its potential to zero (same as
at infinity).
c) How do your answers to a) and b) change?
d) Find the potential difference V between the sphere and the shell in terms of charge Q.
e) What is the capacitance of the system with the sphere and the shell?
Sol: a) Due to the properties of the conductor, especially the fact that the electric field inside the
conductor vanishes and the charges will arrange itself to the surfaces of the conductor, we can read off
the charge densities.
σR =
Q
,
4πR2
σa = −
Q
,
4πa2
σb =
Q+q
.
4πb2
b) Potential inside the conductor is constant and that of a charge located at the origin with charge Q0
0
1 Q
is given by 4π
. Thus we can get the potential as
0 r
φ=













Potential
1 Q+q
φb<r = 4π
r + c4 ,
0
φa<r<b = c3 ,
1 Q
φR<r<a = 4π
+ c2 ,
0 r
φr<R = c1 ,
Constant
c4 = 0
1 Q+q
c3 = 4π
b
0
Q+q
Q
1
(
c2 = 4π
b − a)
0
Q+q
Q
1
c1 = 4π0 ( b − a + Q
R)
B.C.
φ(r = ∞) = 0
φb<r (r = b) = φa<r<b (r = b)
φa<r<b (r = a) = φR<r<a (r = a)
φR<r<a (r = R) = φr<R (r = R)
Q
Q
1
( Q+q
Thus the potential at the center is φr=0 = 4π
b − a + R ).
0
If we connect the outer surface to a grounding wire, we drain the charge of that surface.
c) The charge σb = 0, while the charge densities at r = R, a are not modified. The potential at the
center will be modified in such a way that the potential φr=b = 0 at r = b and φb<r = 0. Thus we get
Q
1
(− Q
φr=0 = 4π
a + R ).
0
d) The potential difference V is
1
1
1
V = φr<R (r = R) − φR<r<a (r = a) =
−
Q.
4π0 R a
e) This the capacitance is given by
1
1
Q
1
−
,
Q=
V =
4π0 R a
C
−→
C = 4π0
aR
.
a−R