Download Exam 1 Solutions

PHY2054 Fall 2014
Prof. Paul Avery
Prof. Andrey Korytov
Sep. 26, 2014
Exam 1 Solutions
1. Charges are arranged on an equilateral triangle of side 5 cm as shown in the
diagram. Given that q1 = 5 µC and q2 = q3 = −2 µC find the magnitude of the
net force on charge q1 (in N).
Answer: 62
q1
q2
q3
Solution: The x component of the force cancels and the y components of the
force on q1 are the same for both charges. Let a be the length of each side. Then the total force in
qq
the y direction is 2 1 2 k sin 60 = 62 N .
a2
2. Charges are arranged on a square of side d as shown in the diagram. In
what direction does the electric field at the center of the square point? (The
quadrants are numbered counterclockwise starting from the positive x-axis.)
Answer: Fourth quadrant
Solution: The electric fields at the origin due to the left & right (+Q and –Q)
and top & bottom (+Q and –2Q) charges do not cancel. The electric fields
due to these two charge pairs are then in the directions +x (for the left & right charges) and –y
(for the top & bottom charges), leading to a net electric field in the 4th quadrant.
3. Two charged particles are fixed to the x-axis: particle 1 of charge q1 = 20 µC at x = 0 m, and
particle 2 of charge q2 = −80 µC at x = 0.6 m. At what coordinate along the x-axis is the net
electric field produced by the particles equal to zero?
Answer: −0.6 m
Solution: Between x = 0 and x = 0.6 m, electric fields due to charges q1 and q2 point in the same
direction and cannot cancel. For x > 0, the two fields are in opposite directions, but the larger in
magnitude charge q2 is closer and hence its field is always greater than the field due to charge q1;
hence, the two fields cannot cancel each other here either. For x < 0, the fields are again in the
opposite directions and would cancel, if their magnitudes are the same:
q1
x2
q2
k=
( L − x)
Solving yields x = −L = −0.6 m.
1
2
k.
PHY2054 Fall 2014
4. An electron with velocity v0 along the +x direction enters a region with a uniform electric field
of magnitude 5000 V/m in the −y direction. If the electron travels 5.0 cm in the x direction and
gets deflected by 1.0 cm in the y direction what is v0 (in m/s)?
Answer: 1.05 × 107
Solution: The deflection by a constant acceleration is y = 12 at 2 , where y = 0.05 m, the
acceleration a = Ee / me and the time t = Δx / v0 . Solving yields v0 = 1.05 × 107 m/s.
5. A conducting sphere with charge +10 nC is placed at the center of two
concentric conducting spherical shells of radius r1 = 2 cm and r2 = 5 cm
(measured to the outer surfaces). The inner shell carries a charge of −7 nC
and the outer shell has a charge of +6 nC. Find the charge (in nC) on the
inner surface of the outer shell.
Answer: −3
Solution: The E field inside the conductor of the outer shell must be 0. A Gaussian surface
drawn within the outer shell will thus have no flux going through it, implying, from Gauss’ law,
that there is 0 charge enclosed by the Gaussian surface. Since the total charge of the central
sphere and inner shell is +3 nC, the charge on the inner surface of the outer shell must therefore
be −3 nC.
6. In the previous problem, find the magnitude of the electric field at r = 50 cm.
Answer: 324 V/m
Solution: Outside the uniformly charged shells/spheres, the electric field is kq / r 2 and points
radially. In this problem, the total field due to the two shells and the central sphere is then
E = kq1 / r 2 + kq2 / r 2 + kq3 / r 2 =
where Q = +9 nC. This yields E = 324 V/m/
2
q1 + q2 + q3
r
2
k=
Q
r2
k,
PHY2054 Fall 2014
7. Four charges of magnitude Q = 3.0 µC (but different signs, as in the figure)
are arranged on the corners of a square of side 25 cm. Find the potential energy
of the system of the four charges (in J).
+Q
-Q
-Q
+Q
Answer: −0.84
Solution: For a system of point charges, the total potential energy is
U=
∑
all (i,j) pairs
kqi q j
r
With four charges, there are 6 pairs: 12, 13, 14, 23, 24, 34. Let L be the length of a side. Four
pairs (corresponding to the 4 sides) each contribute −kQ 2 / L while the diagonals each
contribute kQ 2 / 2L . The total potential energy is thus
⎛ kQ 2 ⎞
⎛ kQ 2 ⎞ kQ 2
4⎜ −
+
2
−4 + 2 = −0.84 J .
⎟
⎜
⎟=
L ⎠
L
⎝
⎝ 2L ⎠
(
)
8. Electrons in a particle beam have a kinetic energy of 3.2 × 10−17 J. What is the magnitude of
the electric field (in V/m) that will stop these electrons in a distance of 0.1 m?
Answer: 2000
Solution: The distance d can be calculated from K = Fd = Eed , yielding E = 2000 V/m.
9. The movement of a charge in an electric field from one point to another at constant speed
without the expenditure of work by or against the field
Answer: none of these
Solution: The work is zero only when moving along an equipotential surface.
10. Two particle each with charge Q are fixed at the vertices of an equilateral
triangle with sides of length a. The work required to move a particle with a charge
q from the other vertex to the center of the line joining the fixed charges is
Answer: 2kQq / a
a
a
a
Solution: The work required of an external force to move the charge q equals to
the change in the potential energy of that charge:
! Qq $ ! Qq $
Qq
Wext = ΔU = U f −U i = 2 #
k & − 2#
k& = 2
k.
a
" (0.5a) % " a %
11. Two isolated conducting spheres are separated by a large distance. Sphere 1 has a radius of R
and an initial charge 3Q while sphere 2 has a radius of 3R and an initial charge 7Q. A very thin
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PHY2054 Fall 2014
copper wire is now connected to the spheres to allow charge to flow between the spheres. How
much charge will be transferred from sphere 2 to sphere 1? (Note that the charge transferred can
be positive, negative or zero.)
Answer: −Q / 2
Solution: After reaching equilibrium, the two spheres must be at the same potential, thus
Q1
Q
k= 2k
R
3R
Let ΔQ be the charge taken from sphere 2 and moved to sphere 1. Then:
3Q + ΔQ
7Q − ΔQ
k=
k
R
3R
from where we obtain ΔQ = −Q / 2 .
12. Four protons are placed at rest on the vertices of a square of side 2.0 µm. The protons are
released simultaneously. When the protons are very far away, what is their speed (in m/s)?
Answer: 430
Solution: Conservation of total energy yields 0 + U i = K f + 0 , where the initial potential energy
(
)( )
K = 4 ( mv ) . From here, we obtain v = ( ke
U i = ke2 / L 4 + 2 (see problem 7) and the final kinetic energy of four protons
1
2
2
2
)(
)
/ mL 2 + 1/ 2 = 432 m/s .
13. Two equipotential surfaces lying near the middle of the space between the plates of a
parallel-plate capacitor are 2.0 mm apart and have a potential difference of 0.0012 volt. The area
of each plate is 7.5 cm2. What is the magnitude of the charge on each plate, (in units of 10−15 C)?
Answer: 4
Solution: The electric field between two uniformly charged plates with charges Q and −Q is
E = σ / ε 0 = Q / ε 0 A , where A is the area of each plate. We calculate E = ΔV / d = 0.6 V/m.
Solving for Q yields Q = 4.0 × 10−15 C .
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PHY2054 Fall 2014
14. A certain parallel plate capacitor with capacitance 12 µF is connected to a source of EMF
with potential 3 V. A dielectric material with κ = 4 is then inserted between the plates of the
capacitor with the capacitor still connected to the circuit. By how much does the energy stored in
the capacitor change?
Answer: 1.6 × 10−4 J
Solution: The energy in a capacitor is U = 12 CV 2 , with V the voltage across it. Since the
capacitor remains connected to the emf source when the dielectric is inserted, the change in
energy is caused only by the change in capacitance ΔC = C f − Ci = κ C − C = κ −1 C . Thus the
(
(
)
)
change in energy is ΔU = 0.5 × 3× 12 × 10−6 × 32 = 162 µJ .
15. An air-filled parallel-plate capacitor has a capacitance of 2 pF. The plate separation is then
doubled and a wax dielectric is inserted, completely filling the space between the plates. As a
result, the capacitance becomes 4 pF. The dielectric constant of the wax is:
Answer: 4.0
Solution: The capacitance of a parallel-plate capacitor is
C = κε 0
A
d
Without the dielectric, the doubling of the plate spacing would change the capacitance from 2 pF
to 1 pF. Thus to achieve 4 pF, the dielectric constant must be 4.0.
16. A parallel plate capacitor with a capacitance of 2.0 nF is charged to have 0.8 µC on each
plate. How much work must be done by an external agent to double the plate separation while
keeping the charge constant?
Answer: +1.6 × 10−4 J
Solution: If the plate separation is doubled, Cnew = C0 / 2 . With the charge unchanged, the
change in energy is thus
Q2
Q2
Q2
ΔU =
−
=
= +160 µJ .
2Cnew 2C0 2C0
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PHY2054 Fall 2014
17. A 9 V battery is connected to a 3 Ω resistor. How much charge passes through the resistor in
3 hours?
Answer: 32,400 C
Solution: The total charge Q = i × t , where i = 3 A and t = 10,800 s. Thus Q = 32,400 C.
18. You have two resistors R1 and R2, made from the same material, with the length and diameter
of R1 both double that of R2. If R1 and R2 are connected in parallel to a battery, find the ratio of
the currents running through R1 and R2 (find I1 / I2).
Answer: 2
Solution: The resistance of a cylindrical resistor is R =
ρL
ρL
=
. The cross sectional area
A π (d / 2)2
of R1 is 4 times that of R2 while its length is double, therefore R1 = 12 R2 . The current in R1 is thus
twice the current in R2.
19. In the circuit diagram shown find the voltage drop across the 3.0 Ω
resistor.
6.0 Ω
12 V
Answer: 4.5 V
12 Ω
4.0 Ω
Solution: There is 12 V applied across the bottom branch, whose total
5.0 Ω
3.0 Ω
resistance is 3 + 5 = 8 Ω (resistors connected in series). Thus the
current is 12 V / 8 Ω = 1.5 A and the voltage drop across the 3 Ω resistor is 1.5 A × 3 Ω = 4.5 V.
20. In the circuit shown find the total energy (in µJ) stored on all the
capacitors. Where C1 = 2 µF, C2 = 4 µF, C3 = 9 µF, C4 = 1 µF, C5 = 3
µF, and V = 10V.
Answer: 182
C2
V
C1
C3
C4
Solution: The total energy can be calculated using the equivalent capacitance and the emf since
from the battery’s perspective the equivalent capacitance behaves in exactly the same way as the
original capacitors. C4 and C5 combine to give C45 = 3.0 µF. Then C2, C45, C3 in series give 1.64
µF. Adding in C1 in parallel then gives a total capacitance of 3.64 µF. The energy is then
U = 12 CV 2 = 12 × 3.64 × 102 = 182 µJ .
6
C5