PHY350 Midterm Exam
Thomas Curtright
14 October 2009
This is a closed book exam.
You may not use your notes. You may not use your textbook.
You may not discuss the exam with anyone except Professor Curtright.
Good luck!
“On my honor, I have neither received nor given aid on this exam.”
Signature:
Printed Name:
Student ID Number:
1
Problem 1 (25 pts)
Consider the scalar and vector functions:
−
f (→
r ) = xy
→
− →
A (−
r ) = y2 x
+ 2xy + z 2 y + 2yz z
Solution: We compute the following:
→
−
∇f
= yx
+ x y
∇2 f
= 0
→
− →
−
∇·A
→
− →
−
∇×A
= 2x + 2y
= 0
→
− →
−
∇· f A
= y 3 + 4x2 y + xz 2 + 2xy 2
→
−
∇2 f A
= 6xy x
+ 4x2 + 4y 2 + 2xy y + 4xz z Ugh again!!
2
Ugh!
Problem 2 (25 pts)
Find the electric potential V (z) a distance z above the center of a uniformly charged flat circular
disk of radius R lying in the xy-plane. The total charge on the disk is Q. What does your answer
give in the two extreme cases, z R and z R?
What is the electric field at the same point?
Solution: Here σ =
Q
πR2 .
V (z) =
=
=
s=R
σ 2πsds
1
√
4πε0 s=0
s2 + z 2
1
σ 2π
R2 + z 2 − z
4πε0
Q
2 + z2 − z
R
2πε0 R2
σ
R+O
zR 2ε0
Q
z+
V (z) ∼
zR 2πε0 R2
V (z) ∼
→
−
E
Ez
z2
−z
R
R4
R2
+O
2z
z3
= Ez z
Q
∂
= − V (z) =
∂z
2πε0 R2
−z
1− √
=
3
z
+ z2
R2
z
σ
+O
zR 2ε0
R
R2
Q
Ez ∼ =
+O
2
zR
4πε0 z
z4
Ez ∼
Q
+O
4πε0 z
R2
z3
Problem 3 (25 pts)
Find the electric field inside a sphere which carries a charge density proportional to the distance
from the center of the sphere, ρ = kr, for some constant k.
If the sphere has a finite radius R, what is the total charge Q? (Express Q in terms of k and
R.) What is the electric field outside the sphere?
→
−
Solution: Obviously, E = E (r) r
Q (r ≤ R) =
r
0
Q (r > R) =
R
0
4πkr r2 dr = πkr4
4πkr r2 dr = πkR4 = Qtot
Gauss’s integral law for the flux then gives
E (r ≤ R) =
E (r > R) =
πkr4
1
=
kr2
4πε0 r2
4ε0
πkR4
kR4
Qtot
=
=
2
2
4πε0 r
4ε0 r
4πε0 r2
Although you were not asked to do so, we plot E (r).
E/const
1.0
0.8
0.6
0.4
0.2
0.0
0
1
2
3
4
E/const versus r/R where const = kR2 / (4ε0 )
4
5
r/R
Problem 4 (25 pts)
Find the electric potential V (r) for the spherical distribution of charge in Problem 3, for r > R.
What is V (r) for r ≤ R?
Solution:
V (r > R) = −
r
∞r
Qtot
dr
4πε0 r2
r
1
Qtot
dr
−
2
4πε0
∞ r
= −
=
∞
Alternatively,
V (r > R) =
V (r ≤ R) = −
=
=
=
Alternatively, from k = Qtot / πR4 ,
r
∞
R
=
Qtot 1
4πε0 r
kR4 1
πkR4 1
=
4πε0 r
4ε0 r
E (r) dr
r
E (r > R) dr −
E (r ≤ R) dr
∞
R
r
1
k
r2 dr
V (R) −
4ε0 R
1
Qtot 1
−
k r 3 − R3
4πε0 R 12ε0
1
kR4 1
1
+
kR3 −
kr3
4ε0 R 12ε0
12ε0
1
1
kR3 −
kr3
3ε0
12ε0
= −
=
E (r > R) dr
V (r ≤ R) =
Qtot
3πε0 R
1−
1 r3
4 R3
OK, to sum up:
kR3 1
4ε0 r/R
kR3 4 1 r3
−
V (r ≤ R) =
4ε0 3 3 R3
2d
2 d 3
−1
d
r dr V (r ≤ R) = 12ε
r dr r = − ε10 kr = − ε10 ρ (r) as it
k r12 dr
0
V (r > R) =
Check: ∇2 V (r ≤ R) =
should.
We also plot V (r).
1 d
r 2 dr
5
V/const' 1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0
1
2
3
4
5
r/R
V /const versus r/R where const = kR3 / (4ε0 ).
Problem 5 (10 Bonus points!)
A hemispherical bowl of radius 4a is made of conducting metal foil (very thin!) and is placed
concentric with a smaller conducting sphere of radius 3a. So, the centers of the hemisphere and
sphere coincide. Put charge Q on the smaller sphere, and charge −Q on the hemisphere. (Note
that the charges will be distributed angularly in a non-uniform way.)
What is the potential at the point on the smaller sphere which is farthest from the edge of the
hemisphere? (Note this point is equidistant from the edge, at distance d = 5a.)
Solution: The potential at the center of the small sphere is just
V =
1
4πε0
Q
Q
−
3a 4a
=
Q
48πε0 a
since the charges Q and −Q are all at distances 3a and 4a, respectively, even though the charges are
distributed angularly in some unknown way. But the potential at the center of the small spherical
conductor is the same as all points on its surface. (It’s a conductor!) Thus, the potential at the
point in question is also given by this expression for V .
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