Reitze/Kumar
PHY2054
Exam 1
February 16, 2011
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Useful Constants:
ke= 8.99×109 Nm2/C2
ε0= 8.85×10-12 C2/(Nm2)
electron charge = -1.6×10-19 C electron mass= 9.11×10-31 kg
V=Volt N=Newton J=Joule m=Meter C=Coulomb g=9.8 m/s2
"milli-" = 10-3 "micro-" = 10-6 "nano-"=10-9 "pico="=10-12
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1. If the electric field has a magnitude of 4 N/C at a distance of 1 meter from an isolated
point charge, at what distance (in m) is the magnitude of the electric field equal to (0.5, 6,
10) N/C?
E = kq/r2. The information given here leads to kq = 4. Thus solving for r when E = (0.5,
6, 10) we get r2= (8, 4/6. 0.4).
(1) (2.83, 0.82, 0.63
2. An electron traveling north enters a region where the electric field is uniform and
points north. The electron will then:
Because the charge of the electron is negative, the force on it, F = qE is pointed
backwards and the electorn will slow down.
(1) slow down (2) speed up (3) veer east (4) veer west
(5) continue with the same speed in the same direction
3. Two charged particles are arranged as shown in the figure with the
+2 C charge on the left and the -4 C charge on the right. In which
region could a third particle, with charge +1 C, be placed so that the
net electrostatic force on it is zero?
Region 1 +2C
Region 2
-4C
It can be region I only. The electric field due to the charge +2C is pointed away from the
charge while the one due to -4C is pointed towards it. In region 2, the two electric fields
are in the same direction and get added. In region 3 they cancel but the larger charge
being closer, always dominates. Only in region 1 can they cancel to yield a zero electric
field.
(1) Region 1 only (2) Region 1 and 2 only (3) Region 3 only (4) Region 1 and 3 only
(5) Region 2 only
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Region 3
4. A 1 gram particle with a charge of 1 milliC starts from rest in a uniform electric field
with magnitude. If the particle moves (5, 30, 50) m in 2 seconds, what is the magnitude
of the electric field (in N/C)?
Motion under constant acceleration (electric field) is described by x = qEt2/2m. Sokve
for E (do not forget that mass is used in kg).
(1) (2.5, 15, 25)
5. Two point particles, one with charge +8.0 ×10-9 C and the other with charge -2.0×10-9
C, are separated by (4, 6, 8) m. What is the magnitude of the electric field (in N/C)
midway between them?
The fields due to the two charges add since in the region in between, they are pointed in
the same direction.
E = 9x109 {8x10-9/4+ 2x10-9/4} = 22.5
The answers on the exam are using a smaller value of the constant k = 8.99x10.9
(1) (22.5, 10.0, 5.625)
6. Two equal and opposite point charges +Q and –Q are located on the y-axis at y = d/2
and y = -d/2 as shown in the figure? What is the magnitude of the electric field at the
point P on the x-axis a distance x = d/2 from the origin? Note ke = 1/(4πε0).
The charges are at a distance d/√2from the point P. The electric field at point P due to
and
. The field die to
the positive charge has
the negative charge has
and
. The x
components cancel but the y components add, yielding the coefficient 2√2 = 2.83.
(1) 2.83keQ/d2
7. In the previous problem, what is the direction of the electric field at the point P on the
x-axis a distance x = d/2 from the origin?
See above.
(1) negative y-direction
8. Four point charges form a square with sides of length L as shown in the figure. Three
of the charges have charge +Q and one has charge –Q. What is the amount of work that
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must be done (against the electric force) to bring in another +Q charge from infinity and
place it at the center of the square?
+Q
The work done = change in the potential energy = 2√2 keQ2/L. A variant on
this problem, would be to calculate the work done in bringing the charge on the
north-west corner from infinity to its present location (answer 0.71 keQ2/L).
(1) 2.83keQ2/L
9. A uniform electric field of intensity (0.2, 0.3, 0.6) N/C is pointing
along the x-axis. If the electric flux through a circular plane with radius
R = 2 m is 2 Nm2/C, what angle θ (in degrees) does the normal of the
circle make with the x-axis if the normal lies in the xy-plane as shown
in the figure?
L
-Q
L
+Q
+Q
y-axis
Normal
Electric Field
θ
x-axis
The flux Φ = EAcosθ. Thus cosθ = Φ/πR2E = 2/πx4x0.2 = 0.796
And θ = 37.3
Circle
(1) (37.3, 58, 74.6)
10. Which of the graphs in the figure represents the magnitude of the electric field as a
function of the distance from the center of a solid charged conducting sphere of radius R?
The critical word here is conducting. Electric field inside a conductor is zero.
(1) E
11. Which of the following is equivalent to 1 Joule of energy?
Imagine the expression for energy in a capacitor. E = Q2/2C = ½CV2. The only other
correct expression would be 1 Farad-Volt2 but it is not an option. (C here stand for
Coulomb).
(1) 1 C2/F
12. When two capacitors are connected in parallel their equivalent capacitance is 9 µF.
When the same two capacitors are connected in series their equivalent capacitance is 2
µF. What is the capacitance (in µF) of the smaller of the two capacitors?
Suppose that the capacitors are C1 and C2. Then C1 + C2 = 9 and C1C2/(C1+C2) = 2 or
C1C2 = 18. Substituting etc. gets you to C12 -9C1 +18 = 0. The two solutions of this
quadratic equation are C1 = 6 and 3, the smaller of which is 3.
(1) 3
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13. If both the plate area and the plate separation of a parallel-plate capacitor are doubled,
the capacitance is:
For a parallel plate capacitor, C = εoA/d. The answer follows.
(1) unchanged
14. A battery with an internal resistance of (0.5, 1.0, 1.5) Ω is connected to a 10 Ω
resistor. If the voltage across the battery (the terminal voltage) is 20 Volts, what is the
EMF of the battery (in Volts)?
E = V + Ir = I(R+r) = V(R+r)/R
(1) (21, 22, 23)
15. The potential difference between point A and B shown in the
figure is (5, 10, 20) V. How much power (in W) is dissipated in the 4
Ω resistor?
Imagine therefore a 5 V battery between points A and B: The equivalent resistance of the
entire circuit is 3+4/3 = 13/3 = 4.33Ω. The battery provides a current I = 5/13/3 =
15/13 A. The potential difference across the 3 Ω resistor is 3x15/13 and the one across
the 4 Ω resistor is 5-45/13 = 20/13. Finally the power dissipated in the 4 Ω resistor is
V2/R = (20/13)2/4 = 0.5917 W.
(1) (0.6, 2.4, 9.5)
16. What is the current in the (5.0, 3.0, 4.0) Ω resistor in the circuit shown
in the figure?
If this problem is considered slowly, there is a reward in having to do less
work. The equivalent resistance of the top branch is (4+4=) 8Ω same as the bottom
branch. The total resistance is 4Ω. The total current from the battery is then 3A which
splits equally in two parts each 1.5A through the top and bottom branches.
(1) (1.5, 1.5, 1.5) A
17. If the switch is closed at t = 0. How much charge is on the capacitor C
(in µC) at t = 10 ms? Assume that ε = (10, 15, 3) V, R1 = 100 Ω, R2 = 500
Ω, and C = 10 µF.
The time constant RC = 600x10-5 = 6 ms. The charging is Q = εC[1-exp(10/6)]. The answers (in µC) follow..
(1) (81.1, 121.7, 24.3)
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18. A copper cable with resistivity ρ = 1.7×10-8 Ω/m is designed to carry a current of
(200, 400, 600) Amps with a power loss of 2.0 W per meter. What is the required radius
of this cable (in cm)?
W = I2R = I2ρl/A. Note that the power loss figure here is given in W/l = 2 W/m.
Therefore A = πr2 = I2ρl/W and leads to the answers.
(1) (1.04, 2.08, 3.12)
19. Suppose the electric company charges 10 cents per kW·h. How much does it cost to
use a 125 W lamp (4, 6, 8) hours a day for 30 days?
Total energy used = 125x4x30 = 15 kW-hour. It follows that the bill is $1.50.
(1) ($1.50, $2.25, $3)
20. A solid conducting shell with inner radius R and outer radius 2R has a net
charge of (+2, +1, +3) C placed on it. If a +1 C point charge is placed at the
center as shown in the figure, how much charge is located on the outer surface
(radius 2R) of the conductor?
conductor
+1 C
R
Gauss’ theorem says that the total flux off of a closed surface is charge enclosed
Since in a conductor the electric field is zero, as would be the flux though a closed
surface entirely in the shell, the charge on the inner surface should be -1C. The charge
on the outer surface (to ensure that the total charge on the shell is 2C) must be 3C.
(1) (+3, +2, +4) C
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2R