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Name:_______________________ ___
PHY2061
9-28-06
Exam 1 Solutions
y
+4q
+2q
−3q
−2q
−2q
60°
x
s
+2q
+2q
1. A central particle of charge −3q is surrounded by a hexagonal array of other charged
particles (q>0). The length of a side is s, and charges are placed at each corner.
(a) [6 points] Find the component of the force along the x-axis (Fx) on the central
particle.
The force along the horizontal direction and along the direction from bottom left to
upper right cancels. Thus, only the force from the upper left charge and bottom right
charges matter. The upper left charge is greatest in magnitude, and the force is
attractive since the middle charge is negative. Thus:
Fx = − K
( 4q )( 3q ) cos 60° + K ( 2q )( 3q ) cos 60°
s2
6q 2 1
q2
Fx = − K 2 = −3K 2
s 2
s
s2
(b) [6 points] Find the component of the force along the y-axis (Fy) on the central
particle.
Fx = + K
Fx = + K
( 4q )( 3q ) sin 60° − K ( 2q )( 3q ) sin 60°
s2
s2
6q 2 3
q2
=
+
3
3
K
s2 2
s2
Page 1 of 9
Name:_______________________ ___
PHY2061
9-28-06
y
+λ
−λ
R
x
2. Consider electric charge distributed along a one-dimensional path in the form shown
as two sections of ¼ of a circle each. The circle is centered at the origin with a radius
of R, and the linear charge density is +λ in the left quadrant and −λ in the right.
(a) [6 points] Find the component of the electric field along the x-axis (Ex) at the
origin (0,0).
The field contribution from the left arc equals that of the right arc. So we can
calculate twice the x-component of the field from the right arc:
π /2
dq
λ Rdθ
cos θ = 2∫ K
cos θ
2
0
R
R2
λ
2K λ
π /2
Ex = 2 K sin θ 0 =
R
R
Ex = 2∫ K
(b) [6 points] Find the component of the electric field along the y-axis (Ey) at the
origin (0,0).
The field cancels in the y direction.
Page 2 of 9
Name:_______________________ ___
PHY2061
9-28-06
y
x
z
3. Consider a cube with side length s = 2 m and one corner at the origin (0,0,0) as
shown.
(a) [6 points] What is the total charge enclosed by the cube if the electric field is
E = 3x 2 ˆi + 5ˆj − 2 z kˆ N/C ?
(
Φ=
qenc
ε0
)
= ∑ i =1 Φ i = Φ ( z = s ) + Φ ( x = s ) others equal 0 because field in those directions is cons
6
= 3s s − 2 ss 2 = s 3 ( 3s − 2 )
2 2
qenc = ε 0 s 3 ( 3s − 2 ) = 32ε 0 = 2.8 × 10−10 C
(b) [6 points] What is the electric charge density (C/m3) at the center of the right
face at x = 2 m if the electric field is the same as in part (a)?
⎛ ∂Ex ∂E y ∂Ez ⎞
−11
3
+
+
⎟ = ε 0 ( 6 x + 0 − 2 ) x = s = ε 0 ( 6 s − 2 ) = 8.85 × 10 C/m
∂y
∂z ⎠
⎝ ∂x
By the way, we could integrate this to get the answer for part (a)
ρ = ε 0∇ ⋅ E = ε 0 ⎜
qenc = ∫ ρ dV = ε 0 ∫ dy ∫ dz ∫ dx ( 6 x − 2 ) = ε 0 s 2 ( 3s 2 − 2s )
s
s
s
0
0
0
qenc = ε 0 s ( 3s − 2 )
3
Page 3 of 9
Name:_______________________ ___
PHY2061
9-28-06
z
R
4. [8 points] A flat nonconducting surface infinite in extent carries a uniform charge
density of σ = 3 × 10−9 C/m 2 . A small circular hole of radius R = 1.5 m has been cut
in the middle of the sheet as shown. Calculate the electric field at a point z = 5 m
away from the center of the hole along an axis perpendicular to the surface. (In other
words, consider z R , but don’t set R / z exactly equal to zero. You may find the
superposition principle useful.)
Using the superposition principle, add the field due to an infinite sheet of charge
(without a hole) plus a small disk of negative charge (same magnitude but opposite
sign charge density).
E = Esheet + E hole
Esheet =
σ
zˆ
2ε 0
q
zˆ when viewed far away ( z R ) and where q = (π R 2 ) ( −σ )
2
z
1 π R 2σ
σ ⎛ R2 ⎞
ˆ
=−
z
=
−
⎜
⎟
4πε 0 z 2
2ε 0 ⎝ 2 z 2 ⎠
Ehole = K
Ehole
E=
R2 ⎞
σ ⎛
σ
1
−
zˆ = 0.955
zˆ = 162zˆ N/C
⎜
2 ⎟
2ε 0 ⎝ 2 z ⎠
2ε 0
Page 4 of 9
Name:_______________________ ___
PHY2061
9-28-06
v
+
+
+
+
+
+
+
5. [8 points] An electron is launched away from the surface of an infinite
nonconducting sheet of charge with a velocity of v = 2 ×106 m/s on a trajectory
perpendicular to the surface. The charge density of the sheet is σ = +5 nC/m2. Is the
electron able to reach a distance infinitely far away from the charged sheet, and if not,
how far does it travel before turning around? The charge of the electron is
q = −e = −1.6 × 10−19 C , and the electron mass is me = 9.11×10−31 kg .
Because the electron is negatively charged and the sheet positive, the electron is
attracted to the sheet by a force
F = qE, where E =
σ
xˆ taking x in the direction of the particle (left) .
2ε 0
Thus, it will stop and turn around.
The difference in electric potential from the surface of the sheet to position x is:
ΔV = −
σ
x
2ε 0
⎛ σ ⎞
⇒ ΔU = qΔV = ( −e ) ⎜ −
⎟x
⎝ 2ε 0 ⎠
At the turn-around
ΔU = K initial
⇒
eσ
1
x = mv 2
2ε 0
2
ε 0 mv 2
z=
= 0.04 m
eσ
Page 5 of 9
Name:_______________________ ___
PHY2061
9-28-06
6. A conducting sphere of radius R1 contains a charge Q. It is surrounded by a
concentric spherical conducting shell of radius R2 > R1 and charge −Q.
(a) [6 points] What is the difference in electric potential between the shell and
the sphere?
-Q
E
Q
E=K
Q
rˆ
r2
+
R1 < r < R2 by Gauss' L:aw and symmetry
ΔV = − ∫ E ⋅ ds
−
ΔV = − ∫
R2
R1
E ⋅ ds = − | E | ds =| E | dr
Q
Q
K 2 dr = − K
r
r
R2
R1
⎛ 1 1 ⎞
= KQ ⎜ − ⎟
⎝ R1 R2 ⎠
(b) [6 points] What is the capacitance of the arrangement of conductors?
Q = C ΔV
1
1
Q=
ΔV
K⎛ 1 1 ⎞
⎜ − ⎟
⎝ R1 R2 ⎠
R1 R2
C=
K ( R2 − R1 )
Page 6 of 9
PHY2061
9-28-06
Name:_______________________ ___
7. [6 points] Initially two electrons are fixed in place with a separation of 2.15 µm. How
much work must be done to bring a third electron in from infinity to complete an
equilateral triangle?
q
r
The work to place a negative charged particle a distance r away from 2 points
charges also of negative charge is the inverse of the change in the potential energy:
Note that for a point charge, the electric potential a distance r away is V = K
e2
W = −ΔU = −qΔV = 2 K = 2.14 × 10−22 J
r
8. [6 points] Sketch the electric field lines for a negatively charged particle above the
surface of a flat perfectly conducting surface (both above and below the surface).
−
conductor
Page 7 of 9
Name:_______________________ ___
PHY2061
9-28-06
9. A 1.5 µF capacitor is charged to a potential difference of 12 V, and the charging
battery is disconnected.
(a) [6 points] What is the energy stored in the capacitor?
q = C1ΔV = (1.5 × 10−6 F ) (12 )
q2
1
2
U=
= C1 ( ΔV ) = 1.08 × 10−4 J
2C1 2
(b) [6 points] If the charged capacitor is then connected in parallel with a second
(initially uncharged) capacitor, and if the potential difference across the first capacitor
subsequently drops to 9 V, what is the capacitance of this second capacitor?
Charge cannot be destroyed, so it remains on the respective plates even though they
are connected to additional plates. But the capacitance must change (capacitors in
parallel), so the potential difference must change to maintain constant q:
Ceq = C1 + C2 (parallel)
ΔV ′ =
C1
q
q
=
=
ΔV
Ceq C1 + C2 C1 + C2
rearranging:
⎛ ΔV
⎞
C2 = C1 ⎜
− 1⎟ = 0.5μ F
⎝ ΔV ′ ⎠
Page 8 of 9
Name:_______________________ ___
PHY2061
9-28-06
10. [6 points] The electric potential along the x-axis (in V) is plotted versus the value of
x, (in cm). Evaluate the x-component of the electrical force (in Newtons, including
sign) on a proton located on the x-axis at x = 10 cm.
The electric field is given by the negative rate of change of the electric potential per
unit length:
ΔV
500 V
=−
= −2500 V/m (from graph)
Δx
0.2 m
F = qE = ( +1.6 × 10−19 C ) ( −2500 V/m )
E=−
so F = −4.0 ×10−16 xˆ N
Force is directed in negative x direction.
11. [6 points] Two wires are made out of the same material (copper). One has a circular
cross section with radius r = 1 mm and a length of 10 cm, the other has a square cross
section with width s = 1 mm and a length of 5 cm. Which wire has the larger
resistance?
R=ρ
L
A
L1
ρ 0.1 ρ 5
=
= 10
2
π r π 10−6 π
L
ρ 0.05 ρ 5 ρ 5
= 10 > 10
square: R2 = ρ 22 =
π 0−6
π
s
2
Square wire has larger resistance.
circular: R1 = ρ
Page 9 of 9