Download Exam 1 Solutions

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PHY2061
9-29-05
Exam 1 Solutions
+2q
7q
5q
+3q
3q
2q
3q
3q
+4q
+4q
d
5q
7q
+2q
1. [6 points] A central particle of charge 2q is surrounded by a square array of charged
particles. The length of the square side is d, and charges are placed at the square
corners, at the midpoint of the sides, and midway between the corner and the
midpoint for two sides. What are the magnitude and direction of the net electrostatic
force on the central particle due to the other particles?
Note that most charges come in pairs with the same sign and at equal and opposite
distances from the central charge. Thus the net force on the central charge cancels.
The only charges which do not balance lie along a horizontal line crossing the
central charge.
Left side:
 3q  2q  xˆ attractive, force to left
F  K
2
 d / 2
Right side:
 3q  2q  xˆ repulsive, force to left
F  K
2
 d / 2
Total:
q2
F  48K 2 xˆ
d
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Name:_______________________ ___
PHY2061
9-29-05
2. [6 points] Particles 1, with a charge q1 , and 2, with a charge q2, are on the x-axis with
particle 1 at x = 2a and particle 2 at x = a. For the net force on a third charged
particle placed at the origin to be zero, what must be q2 in terms of q1 ?
q2
q1
x= a
x=2a
The only way for the forces to balance is if q1 and q2 have the same sign charge, and the
charge placed at the origin has the opposite sign. Let’s assume that q1 and q2 are both
positive, and q3 is negative. Then the forces on the third particle are:
| q3 q2 |
xˆ
a2
|q q |
F31  K 3 21 xˆ
4a
F32   K
Requiring these to balance, add to zero, implies that
q2 
q1
4
Page 2 of 11
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PHY2061
9-29-05
y
E

x
3. Consider electric charge distributed along a one-dimensional path in the form shown
as ¾ of a circle. The circle is centered at the origin with a radius of R, and the linear
charge density  is positive.
(a) [6 points] Find the component of the electric field along the x-axis (Ex) at the
origin (0,0).
(b) [6 points] Find the component of the electric field along the y-axis (Ey) at the
origin (0,0).
Note that the ¾ ring of positive charge is equivalent to a full ring of positive charge
plus a ¼ ring of negative charge in the first quadrant. For the full ring, by symmetry,
the electric field must cancel. That leaves the field from the negatively charged
segment, which would point at an angle of 45 (that is Ex and Ey are equal and both
positive in the shown coordinate system).
Let’s solve for each component explicitly for a ¼ ring with charge density
  3 C/m :
dq   ds   Rd
dq
cos 
R2
 /2
R
K
K
 /2
Ex   K 2 cos  d 
sin  0 
0
R
R
R
dq
dE y  K 2 sin 
R
 /2
R
K
K
 /2
E y   K 2 sin  d  
cos  0 
0
R
R
R
dEx  K
Indeed the components are equal, and the angle is 45. We could also have integrated
the ¾ positive ring from  / 2    2 and obtained the same answer.
Page 3 of 11
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PHY2061
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4. [8 points] In the figure shown, a small, non-conducting ball of mass m  3 106 kg
and charge q  4.8  C (distributed uniformly through its volume) hangs from an
insulating thread that makes an angle  with a vertical, uniformly charged nonconducting sheet (shown in cross section). The sheet has a surface charge density of
  3 1011 C/m 2 . Considering the gravitational force of the ball and assuming that
the sheet extends far vertically and into and out of the page, calculate the angle .
By the relation of forces,
tan  
FE qE

Fg mg
Now for an infinite sheet of charge, the magnitude of the electric field is E 
So:
4.8 106 C  3 1011 C/m 2 

q
tan  

2 0 mg 2 8.85 1012 C2 / N m 2  3 106 kg  9.8 m/s 2 
tan   0.277    15.5
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
2 0
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y
x
z
5. Consider a cube with side length s = 0.5 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   2xˆ  4 y 3yˆ  6 z zˆ  N/C ?
By Gauss’ Law,
E 

S
Ε  dΑ 
qenc
0
6
 E   Ε  Αi
(sum over flux through all 6 sides)
i 1
Only the y and z components of the electric field vary and thus give rise to a net
flux out of the cube (the constant x component has zero net flux). Moreover, The
face at z=0 and the face at y=0 has E=0, so zero flux as well. That leaves just the
faces at z=s and y=s:
 E  E  A top  E  A front
 E   2xˆ  4 y 3 yˆ  6 z zˆ   s 2 yˆ   2xˆ  4 y 3 yˆ  6 z zˆ   s 2 zˆ
 E  4 s 3 s 2  6 ss 2  4 s 5  6 s 3  0.875
E 
qenc
0
 qenc  0.875  8.85 1012   7.75 10 12 C
(b) [6 points] What is the electric charge density (C/m3) at the center of the top
face at y = 0.5 m if the electric field is the same as in part (a)?
E 

0
E 
Ex E y Ez


 0  12 y 2  6  9 at y  0.5
x
y
z
Gauss' Law in differential form
   0  E  8.0 1011 C/m3
Page 5 of 11
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PHY2061
9-29-05
6.
(a) [6 points] A solid non-conducting sphere of radius R  0.05 m has a uniform charge
density   0.01 C/m3 constant throughout the volume of the sphere. What is the ratio
of the magnitude of the electric field at r  R / 2 to that at r  R ?
Apply Gauss’ Law to a spherical surface with radius r:
q
 E   Ε  dA  enc
0
S
4 r 2 E 
qenc
0
qenc
r2
Now the total charge enclosed is obtained by integration:
4
qenc    dV    r 3
V
3
 qenc  r  R   5.2 C
 E r   K
 qenc  r  R / 2  
1
qenc  r  R   0.65 C
23
So the ratio of the electric field is
E  r  R / 2  Kqenc  r  R / 2 
R2
1 22 1



2
E r  R
Kqenc  r  R  23 1 2
 R / 2
(b) [6 points] Suppose the solid non-conducting sphere in part (a) has a non-uniform
charge density   r   0 r, where 0  0.267 C/m4 (i.e. varies with radius). What is
the ratio of the magnitude of the electric field at r  R / 2 to that at r  R ?
The difference with part (a) is in calculating the enclosed charge:
qenc  r  R     dV    0 r 4 r 2 dr  4 0
R
V
 qenc  r  R   5.2  C
0
R4
  0 R 4
4
1
qenc  r  R   0.33 C
24
So the ratio of the electric field is
E  r  R / 2  Kqenc  r  R / 2 
R2
1 22 1



2
E r  R
Kqenc  r  R  24 1 4
 R / 2
 qenc  r  R / 2  
Page 6 of 11
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PHY2061
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R3
R1
R2
7. Three charged, concentric, spherical conducting shells have radii
R1  10 cm, R2  15 cm, R3  20 cm . The thickness of the conducting shells, while
not zero, is considered negligible. The charge on the innermost shell is 15C, the
charge on the middle shell is 20C, and the charge on the outermost shell is +25C.
(a) [6 points] What is the charge on the inner surface of the outermost conducting shell?
The electric field must be zero inside an isolated conductor. So a spherical Gaussian
surface within each conductive shell must have zero flux since E=0, which implies that by
Gauss’ Law the enclosed charge must be zero. Thus, the inside surface of each shell must
have an electric charge that exactly balances the charge from the inside conductors.
Therefore, the inside surface of the outermost shell must balance a charge of 35C,
which is thus +35C.
(b) [6 points] What is the direction and magnitude of the electric field at radius
r  17 cm ?
The two inner shells have a total electric charge of 35C. By Gauss’ Law, the flux
through a spherical surface with radius 17 cm enclosing these two shells must equal the
enclosed charge / 0:
E 

S
Ε  dA 
qenc
0
35 C
35 C
  9 109 
 1.09 107 N/C
2
4 0 r
r2
The direction of the electric field in this region is inward toward the center of the
sphere(and toward the negative charge)
E
Page 7 of 11
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PHY2061
9-29-05
E
v
p
+
+
+
+
+
+
+
8. [8 points] A proton is injected with a velocity of v  2 106 m/s into a region of
uniform electric field between two large plates separated by 1m and maintained with
an electric potential difference of 30,000 V. The proton travels from lower electric
potential to higher on a path perpendicular to the plates. Does the proton reach the far
plate, and if not, what is the distance of closest approach to it? The charge of the
proton is e  1.6 1019 C , and the proton mass is mp  1.67 1027 kg .
As the proton climbs from lower electric potential to higher, it will gain potential
energy at the expense of kinetic energy. The total potential energy to climb to reach
the other plate is:
U  qV  1.6 1019 C   30000 V   4.8 1015 J
The initial kinetic energy is
K
2
1 2
mv  0.5 1.67 1027 kg  2 106 m/s   3.34 1015 J
2
which is not enough energy to reach the far plate, as the energy is short by
1.46 1015 J  1.6 1019 C   9125 V   9125 eV
For parallel plates, the electric field is uniform and the equipotential lines equally
spaced. So the distance of closest approach to the far plate is:
9125
1.0 m   0.3 m
30000
a little less than 1/3 of the plate separation.
dca 
Page 8 of 11
Name:_______________________ ___
PHY2061
9-29-05
9. [6 points] The electric potential along the x-axis (in kV=103 V ) is plotted versus the
value of x (in meters). Evaluate the x-component of the electrical force (in Newtons)
on a charge of 4.70 C located on the x-axis at x= 1.2 m.
The electric field is given by the negative rate of change of the electric potential per
unit length:
6000   3000  V
V

 7500 V/m (by estimation from graph)
x
1.2 m
F  qE   4.7  106 C   7500 V/m 
E
so F  0.035xˆ N
Force is directed in negative x direction.
Page 9 of 11
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PHY2061
9-29-05
A
B
10.
(a) [6 points] Consider the shown arrangements of capacitors above. Calculate the
effective capacitance, Ceff , of the network between the terminals A and B given that
each of the shown capacitors has a capacitance C = 4 F.
1
C
1 1
The rightmost two capacitors are in series, with capacitance C      
2
C C 
This is in parallel with the left capacitor, so the effective capacitance is:
3
Ceff  C  C   C  6 F
2
A
  
B
(b) [6 points] Consider the above infinite chain of capacitors. Calculate the effective
capacitance, Ceff , of the network between the terminals A and B given that each of
the shown capacitors has a capacitance C = 4 F.
You could solve this problem by successive approximations. Suppose just two
capacitors were added to the end of the chain in part (a). The last two capacitors
would be in series, so 2F capacitance. Adding this in parallel with 4F gives 6F,
in series with 4F again gives 2.4F in parallel with 4F. So the two loop capacitor
case gives 6.4F. Three loops give 6.46F, and so on.
However, this semi-infinite ladder can be solved inductively. Let Ceff be the
capacitance of the circuit. Then the effective capacitance of the circuit is capacitor C
in parallel with a circuit containing a capacitor C in series with Ceff !
Page 10 of 11
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PHY2061
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Ceff
1
1 
 C  

 C Ceff 
1
1
C C 
 Ceff  C   eff

 CCeff 
  Ceff  C  Ceff  C   CCeff
 Ceff 2  CCeff  C 2  0
C  C 2  4C 2
by quadratic formula
2
C
 1  5  1.62C  6.5 F
2
 Ceff 
 Ceff


11. [6 points] A parallel plate capacitor with a capacitance of 12 F is connected to a
battery that maintains an electric potential difference of 6V. If the plates are
squeezed to one-third of the original separation, how much work must be done by the
battery to maintain the same potential difference across the plates?
1
The stored potential energy of a capacitor is U  CV 2 . The capacitance of a
2
A
parallel-plate capacitor satisfies: C   0 , so if the plates get 3 times closer, the
d
capacitance is 3 times larger  36F. Thus the potential energy increases since the
voltage remains the same across the capacitor but the capacitance increases. The net
work done by the battery to change this potential energy is:
1
W  U  U f  U i  V 2  C f  Ci 
2
1
2
U   6 V   36  12 F   4.32  104 J
2
Page 11 of 11