Phy481 Exam 2

NAME _____________________
STUDENT NO. ___________________
Phy481 Exam 2
November 10, 2006
No books, notes, calculators, cell phones, etc.
SHOW ALL WORK
ANSWERS WITHOUT JUSTIFICATION WILL NOT BE CONSIDERED
Solutions of Laplace’s equation
Cartesian coordinates (no z dependence) (boundary conditions at y = ± a 2 )
V (x, y) = X (x)Y ( y) , where Yk ( y) = Ak cos ( kπ y a ) + Bk sin ( kπ y a ) and
X k (x) = Ck e
kπ x a
+ Dk e
− kπ x a
(unbounded at
x = +∞ , C = 0 ; or at x = −∞ ,
D = 0 ) or
X k (x) = Ck cosh kπ x a + Dk sinh kπ x a if bounded at x = ± a 2
A general solution is a sum of solutions for integer k’s allowed by boundary conditions.
Spherical coordinates (no φ dependence)
∞
V (r,θ ) = ∑ R (r)Θ  (θ ) , where R (r) = A r + B r

− ( +1)
=0
and Θ  (θ ) = P ( cosθ ) , the
Legendre polynomials: P0 (cosθ ) = 1, P1 (cosθ ) = cosθ , P2 (cosθ ) = ( 3cosθ − 1) 2,…
Cylindrical coordinates (no z dependence)
∞
(
V (r,φ ) = Aln r + B + ∑ Rn (r)Φ n (φ ) , where Rn (r) = An r + Bn r
n=1
Φ n (φ ) = ( Cn cos nφ + Dn sin nφ ) .
n
−n
) and
NAME _____________________
STUDENT NO. ___________________
1. [20 pts] Fill in the blanks assuming electrostatic equilibrium (2 pts each, no partial
credit).
a) The electric field inside a conductor is zero.
b) The potential inside a conductor is constant
c) Inside a conductor the charge density is zero
d) The charge on a conductor resides only on the surfaces
e) A net charge +Q is always paired with a charge –Q elsewhere.
f) What is the orientation of the electric field at the surface of a conductor?
perpendicular to the surface
g) Relate the charge density on a conductor to the electric field at the surface:
σ = ε0 E
h) How is the electric field E outside an uncharged conducting spherical shell related to
a charge Q at its center?
The field is the same as the field of the charge Q
i) How is the electric field E outside an grounded conducting spherical shell related to
the charge Q at its center?
The field outside is zero.
j) How is the force per unit area on the surface of a conductor related to the surface
charge density?
F = qEremote = q
σ
σ
; where field of remote charges, Eremote =
2ε 0
2ε 0
2
F q σ
σ
=
=
A A 2ε 0 2ε 0
NAME _____________________
STUDENT NO. ___________________
2. [25 pts] Capacitors C1 , and C2 carry charges Q1 and Q2 respectively.
The switch is closed causing the charge to redistribute.
a) [15 pts] What are the new charges Q1′ and Q2′ ?
Charge is conserved, and the final voltages are equal.
Q1′ + Q2′ = Q1 + Q2
Q1′
C1
Q1′ +
C2
C1
=
Q2′
C2
= V ′; Q2′ =
C2
C1
Q1′
Q1′ = Q1 + Q2
Q1′ =
Q1 + Q2
C Q + Q2
Q + Q2
; Q2′ = 2 1
= 1
C2 ⎞
C2 ⎞ ⎛
C1 ⎞
⎛
C1 ⎛
⎜⎝ 1 + C ⎟⎠
⎜⎝ 1 + C ⎟⎠ ⎜⎝ 1 + C ⎟⎠
1
1
2
b) [10 pts] Compare the stored energy before and after the switch is closed. How is this
energy dissipated?
Net charge is conserved, so amount leaving one capacitor, ΔQ , must arrive at the other.
Q1 = Q1′ + ΔQ; Q2 = Q2′ − ΔQ
1 Q1
1 ( Q1′ − ΔQ )
1 Q2
1 ( Q2′ + ΔQ )
U1 =
=
; U2 =
=
2 C1 2
C1
2 C2 2
C2
2
2
2
Q′ΔQ ( ΔQ )
Q′ ΔQ ( ΔQ )
1 Q1′
1 Q2′
U1 + U 2 =
− 1
+
+
+ 2
+
2 C1
C1
2C1
2 C2
C2
2C2
2
2
2
2
2
⎛ Q′ Q′ ⎞ ( ΔQ ) ⎛ 1
1⎞
= U1′ + U 2′ + ΔQ ⎜ 2 − 1 ⎟ +
+
2 ⎜⎝ C1 C2 ⎟⎠
⎝ C2 C1 ⎠
2
Final voltage is the same for both capacitors,
Q1′
C1
=
Q2′
C2
, leaving only the ( ΔQ ) term.
2
( ΔQ ) ⎛ C1 + C2 ⎞
U1′ + U 2′ = U1 + U 2 −
2 ⎜⎝ C1C2 ⎟⎠
2
1 ( ΔQ )
ΔU = −
; where C = C1C2 ( C1 + C2 )
2 C
Changing currents radiate energy and, if the wires have a finite resistance, they heat up.
2
NAME _____________________
STUDENT NO. ___________________
3. [25 pts]
a) [10 pts] A charge q 0 is placed a distance z 0 from the
center of a grounded conducting sphere of radius R. To
determine the potential everywhere, use the image
method to find the image charge q1 and location z1 that
makes potential zero on the sphere.
b) [5] Write the potential everywhere in terms of the given
parameters.
q1
−q0
−q ( R − z1 )
=
; q1 = 0
R − z1 z0 − R
z0 − R
q1
R + z1
=
−q0
z0 + R
; q1 =
( R − z1 ) = ( R + z1 ) ;
z0 − R
z0 + R
2
z q
0
z0
a
z1
−q0 ( R + z1 )
q0
q1
⎤
1 ⎡
+
⎢ 2
⎥
2
2
4πε 0 ⎢ r + z 2 − 2rz cosθ
⎥⎦
r
+
z
−
2rz
cos
θ
⎣
0
0
1
1
R
q1
q0
2
a = z0 , b = R a; q1 = −q0 R a
2
q0
2
q1q0
2
−
q1q0
2
q0
R
F= 2+
+
= 2−
2
2
2
a
( a + b) ( a − b)
4a
4a
a+ R a
4a
R⎛
1
3⎜
a ⎝ 1 + R2 a2
(
q0
+
)
2
2
q0
( a − R a)
2
2
1
⎞
2
;
= 1  2x + 3x + …
2
2
⎟
2
2
1 − R a ⎠ (1 ± x )
) (
2
(
R
−
a
1
)
2
4
2
4
⎞ ⎛
⎞⎤
R ⎡⎛
R
R
R
R
= 2 − 3 ⎢⎜ 1 − 2 2 + 3 4 + …⎟ + ⎜ 1 + 2 2 + 3 4 + …⎟ ⎥
⎠ ⎝
⎠⎦
4a
a ⎣⎝
a
a
a
a
4
⎤
1
2R ⎡
R
1
2R
2
2
=
1
+
3
+
…
≈ 3 assuming R a << 1
⎢
⎥;
2
3
4
2
⎦ 4a
4a
a ⎣
a
a
a
-3
R≈
(next term is ~10 a)
8
1
q1
b
2
1
R − z1
( R − z1 )( z0 + R ) = ( R + z1 )( z0 − R )
c) [10 pts] Shown are two charges q0 separated by
the distance 2a. The magnitudes and locations
of two additional charges q1 are chosen so that
the potential is zero on a sphere of arbitrary
radius R centered on the midpoint (see above).
Find the approximate radius R of a grounded
conducting sphere that will make the force zero
on either charge q0.
0=
⎫
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎭
−R
R
z0 + R
2R = 2z1z0 → a) z1 = R z0 ; q1 = −q0 R z0
b) V (r,θ ) =
⎫
⎪
⎪
⎪
⎬z
⎪ 0
⎪
⎪
⎭
q1
q0
b
a
NAME _____________________
STUDENT NO. ___________________
+V0
–V0
4. [25 pts] Pictured is a long conducting hollow cylinder of radius
r
R
R, with the 1st and 3rd quadrants at potential +V0 and the 2nd
and 4 quadrants at potential −V0
φ
th
a) [5 pts] Using the function f n (φ ) = An sin nφ what value of n +V
0
(integer) makes f n the best approximation to this
f 2 = A2 sin 2φ
+V0
π
4
potential, and plot it on the graph to the right.
n = 2 f n = A2 sin 2φ
zeros: π 2, π , 3π 2, 2π , …
critical points: π 4, 3π 4, 5π 4, 7π 4, …
–V0
π 2 3π
4
–V0
b) [5 pts] Consider the critical points of f n ( φ where f n is
π
+V0
either maximum or minimum) and the zeros of f n ( φ
π
4
where f n (φ ) = 0 ). What is the next integer m , where the
zeros of f m include all the zeros of f n , and where the
f6 = A6 sin 6φ
π 2 3π
4
critical points of f m include all the critical points of f n ?
φ
2π
3π 2
π
–V0
Plot the f m on the graph to the right.
n=6
zeros: 0, π 6, π 3, π 2, 2π 3, 5π 6, π , …
critical points: π 12, π 4, 5π 12, 7π 12, 3π 4, 3π 4, 11π 12, …
c) [15] Determine the series solution for the potential V (r,φ ) that satisfies Laplace’s
equation outside the cylinder. (Hint: to save time double the Fourier integrals between
0 and π ). Determine the first two terms in the series and compare them to the
functions in parts a) and b).
n
Zero net charge: no Aln r or B ; Outside cylinder: no r terms;
Odd boundary condition: no cos nφ .
∞
V ( r,φ ) = ∑ An r
−n
n=1
∞
sin nφ ; Boundary: V ( R,φ ) = ∑ An R sin nφ
−n
n=1
Fourier integrals for V (R,φ ) = ±V0
1
π
2π
∞
0
n=1
∫ V ( R,φ ) sin mφ dφ = ∑ An R
=
−n
2π
∫ sin nφ sin mφ dφ = δ nm
0
{− cos mφ
mπ
2V0
π 2
0
π
+ cos mφ π
2
{−2cos m π 2 + 1 + cos mπ }
} = 2V
mπ
0
φ
NAME _____________________
{
1
π
2π
−2cos m
8V0
STUDENT NO. ___________________
}
π
+ 1 + cos mπ = 0
2
=4
=0
=4
∫ V ( R,φ ) sin mφ dφ = mπ = Am R
−m
0
8V0
∞
→ Am =
8V0 R
mπ
(m = 1)
(m = 2)
(m = 3,4,5)
(m = 6)
m
; m = 4n − 2, n = 1,2,3,…
4n−2
1 R
V ( r,φ ) =
sin [ (4n − 2)φ ]
∑
π n=1 4n − 2 r 4n−2
Dependence of first two terms on φ are: sin 2φ and sin 6φ , as in parts a) & b) above.
NAME _____________________
STUDENT NO. ___________________
5. [25 pts] The figure on the right represents a “line
dipole”, parallel lines of charge with opposite charge
densities λ , separated by a distance 2d.
+
r
r−
a) [10 pts] Use the general solution to Laplace’s
equation in cylindrical coordinates to determine the
φ
potential of a single line of charge with charge
−
λ
density λ . (You may need to find the field and
Gauss’s Law to determine an unknown constant.)
A
d
d
V (r,φ ) = Aln r + B; E = −∇V = − r̂
r
Gauss’s Law on a cylinder of radius r and length L centered on the single line charge:
λL
λ
λ
2π rLEr =
; Er =
→ A= −
ε0
2πε 0 r
2πε 0
λ
ln r + B
2πε 0
At the wire radius, a, let the potential, V (a,φ ) = 0 .
λ
λ
V (a,φ ) = 0 = −
ln a + B; B =
ln a
2πε 0
2πε 0
The potential is then
λ
λ
λ
a
V (r,φ ) = −
ln r +
ln a =
ln
2πε 0
2πε 0
2πε 0 r
V (r,φ ) = −
b) [10 pts] For the two charged lines in the figure find the first term in an expansion for
the potential at any point (r,φ ) where r >> d , and identify in this potential the line
dipole moment pline .
Summing the two potentials at r : Vd (r,φ ) =
2
2
λ
a
−λ
a
λ
ln +
ln =
( ln r− − ln r+ )
2πε 0 r+ 2πε 0 r− 2πε 0
2
(
2
r+ = r − 2rd cos φ + d ; r− = r + 2rd cos φ + d → r± = r 1  2 d cos φ r + d
(
)
(
2
r
1
1
2
2
2
2
ln r+ = ln r + ln 1 − 2 d cos φ r + d r ; ln r− = ln r + ln 1 + 2 d cos φ r + d r
2
2
)
2 12
)
−2
2
Use expansion: ln(1 + x) = x − x 2 + … with x = ±2 d cos φ r ; drop terms r or higher.
λ
2λ d cos φ pline cos φ
Vd (r,φ ) =
ln r− − ln r+ ) ≈
=
, where pline = 2λ d
(
2πε 0
2πε 0 r
2πε 0 r
c) [5 pts] Compare the r dependences of this potential and that of a point charge dipole,
and state why they might be the same or different.
−2
−1
Point dipole ∝ r , while the line dipole ∝ r . Line dipole charge extends to ±∞ .
r+
+λ