Download PHY 2054 Fall 2012 Kumar/Mitselmakher Exam I

PHY 2054
Fall 2012
Kumar/Mitselmakher
Exam I- Solutions
1. On an isosceles triangle with the equal sides 5 m long and a base of length 6 m, the
top vertex contains a charge (8, 32) nC. The two vertices at the base contain a
charge of -4 nC.. Calculate tthe
he magnitude (in N/C) and direction of the electric field
at point P, the midpoint of the base.
Answer (4.5,18), down
Solution: Note here that the electric field due to the lower two negative charges is equal in magnitude
but opposite in direction and
d cancel. The only electric field at P is the one due to the charge at the top
vertex. Also the vertical distance between the top vertex and point P, by Pythogorus theorem is 4 m.
The E-direction
direction then is down and the magnitude
.
2. For the charge distribution described above, calculate the work done (in nJ) in moving another
test charge of 7 nC from far away to the midpoint of the base.
Answer: (42, 336)
Solution: Here we use the relationships (1) W = - ∆U = -q∆V = -qVP since V(∞) = 0. (2)
Thus the work done W = -7x 10-9 x --6 = 42 nC.
3. Consider the three electric charges A ((-q),
q), B (+q) and C (+q). They are arranged collinearly as
a
shown in the figure. Rank these charges in order of increasing magnitude of the net force they
experience.
Answer: CAB
Solution: A gets attracted by both B and C. The forces due to B and C are in the same direction (x) and
they are of different magnitude
tude because of the different distances. B is attracted by A and repelled by C.
There we have forces that are in the same direction ((-x).
x). They are also of the same magnitude. Finally C
is attracted by A (in the -xx direction) but repelled by B (in the x direction) and these forces partly cancel.
Thus C feels the least force, B feels the most and A is in between.
4. A horizontal beam of electrons initially moving at 3.9 107 m/s is deflected vertically up by the
vertical electric field between oppositely charged parallel plates. The magnitude of the field is
2.30 104 N/C. What is the vertical deflection d of the positrons as they leave the plates?
Answer: 0.531 mm
Solution: The electron moves horizontally at speed 3.9x107 m/s without acceleration and vertically
v
with
zero initial speed but acceleration a = eeE/m.
12 . . "
.
.!
5.312 10&'
5. A metal sphere A has charge Q. Two other spheres, B and C, are identical to A except they have
charge -Q . A touches B, then the two spheres are separated. B touches C, then those spheres
are separated. Finally, C touches A and those two spheres are separated. What is the
magnitude of the force between the spheres A and C as a ratio to the force between them in the
beginning?
Answer: 1/16
Solution:
A
B
C
Beginning Charge
Q
-Q
-Q
A touches B
0
0
-Q
The force between A and C in the beginning( )
*
+
B touches C
0
-Q/2
-Q/2
. At the end, it becomes, ( )
C touches A
-Q/4
-Q/2
-Q/4
*
+
(
.
6. We have a hollow metallic sphere with charge -5.0 μC and radius 5.0 cm. We insert a +10 μC
charge at the center of the sphere through a hole in the surface. What charge now rests on the
outer surface of the sphere?
Answer: +5 μC
Solution: Consider Gauss’ theorem. The electric field inside the metal must be zero. That means that
there is a charge of - 10μC on the inner surface of the metal shell. Since the total charge on the metal
must be -5 μC, the residual is +5 μC which must stay on the outside surface.
7. A Van de Graaff generator has a spherical dome of radius 20 cm. Operating in dry air, where
“atmospheric breakdown” is at Emax = 3.0x106 N/C, what is the maximum charge that can be held
on the dome?
Answer: 1.3 x 10-5 C
*
Solution: Since , ) - , it follows that ./0 - 123
4
1.33 10&5C
8. There is a hollow, conducting, uncharged sphere with a negative charge inside the sphere.
Consider the electrical potential at the inner and outer surfaces of the sphere. Which of the
following is true?
Solution: Since the electric field in a conducting medium is zero, the potential is constant. The potential
is the same on the inner and outer surfaces.
9. Using a 1-mF capacitor, a 2-mF capacitor, and a 3-mF capacitor, which of the following
capacitances cannot be made by a combination that uses all three? (Hint: At most only 2
combinations must be considered to determine the correct answer).
Answer: 5/11 mF
Solution: With capacitors, a series combination is the lowest equivalent capacitor and a parallel
combination is the highest one. They are respectively, 6/11 mF and 6 mF. Among the capacitance
values given, only one is outside this range.
10. A capacitor is attached across a battery and charged. Then the battery is removed leaving the
capacitor charged. The positive lead of the capacitor is then connected to one lead of a
previously uncharged identical capacitor, and then the other lead of the charged capacitor is
connected to the other lead of the second capacitor. How does the energy Eo stored in the
originally charged capacitor compare to the energy Ef stored in the connected capacitors?
Answer: Eo = 2Ef
*
Solution: When the capacitor is originally charged, its energy is ,6 .7 . With the battery
8
disconnected, the total charge remains the same when you plug it into another capacitor. The total
capacitance is now 2C. The final energy then becomes, ,9 energy if you had left the battery connected?
*
8
:
. What would be the final
11. A high voltage transmission line of diameter 2 cm and length 200 km carries a steady current of
1 000 A If the conductor is copper with a free charge density of 8 x1028 electrons/m3, how long
does it take one electron to travel the full length of the cable?
Answer: 8x108 s.
Solution: The current density j = I/A = n e v. Here A is the cross sectional area of the wire and v is the
average drift velocity for the electrons. The time taken to travel a distance L, T = L/v = LAne/v = 8.042 x
108 s. This amounts to more than 25 years.
12. How long is a wire made from 100 cm3 of copper if its resistance is 8.5 ohms? The resistivity of
copper is 1.7x 10-8 Ω·m.
Answer: 224 m
Solution: The resistance of a wire R = ρl/A but we also have V = Al. Removing A in favor of v, we get R =
ρ l2/V. Solving for l = √[VR/ρ] = 223.6 m.
13. By what factor is the resistance of a copper wire changed when its temperature is increased from
20°C to 120°C? The temperature coefficient of resistivity for copper = 3.9 x 10-3 (°C)-1.
Answer: 1.39
Solution: The temperature dependent change of resistance is calculated here assuming that there is no
change in the size of the wire (no thermal expansion). In that case:
Or
14. All resistances are 2 Ohms and the emf is 10 V. Calculate
the total current drawn from the emf source.
Answer: 2A.
Solution: Note here that the resistor R1 is ba
basically shunted
through. There are only four resitances in this citcuit; R2 and R3
in parallel (1Ω)) and the combination in series with R and R4. The
total equivalent resistance is 5Ω.. The total current drawn is 10/5
= 2A.
15. What is the potential difference between
points a and b?
Answer: 6V
Recall that in order to calculate the potential difference
between two points, Vab = Va – Vb, you start at b, go to a
and calculate the potential difference followi
following
Kirchhoff’s rules.
But firstt to calculate the currents in different branches, we have noted the two junctions a and b;
associated currents I1 with branch bcda; I2 with ba and I3 with befa. The directions are chosen as shown
in the figure. The junction rule tells us that I3 = I1+I2.
The loop bcdab gives:
16 - 4I1 +12I2 -18 = 0
Or 2I1-6I2 = -1
The loop cdafebc gives:
16 -4I1 -8(I1+I2) = 0
3I1 +2I2 = 4
The answer is I1 = 1A and I2 = 1/2 A.
It follows that Vab = 18-12
12 x 0.5 = 12V