Download KE q V 1.6 10 50 100 8 10 = Δ = − × − = × KE 8 10 J = ×

Key
Name __________________________
Phys 104: College Physics
MULTIPLE CHOICE
1)4 How much kinetic energy will an electron need to get from point A (at a potential of +100 V)
to point B (at a potential of +50 V) in the figure below?
A 100V
a) none
−19
−18
KE = qΔV = −1.6 × 10
50 − 100 = 8 × 10
b) 50 J
c) 8.0 x 10-18 J
KE = 8 × 10 −18 J
B 50V
-17
d) 2.4 x 10 J
e) 1.14 x 10-27J
(
)(
2)4 a) When resistors (R1 ≠ R2) are in series, the
i) current flowing through each is the same.
ii) the voltage across each is the same.
b) When resistors (R3 ≠ R4) are in parallel, the
i) current flowing through each is the same.
ii) the voltage across each is the same.
c) Capacitors 1 & 2
i) are at the same voltage.
ii) have the same charge.
d) Capacitors 3 & 4
i) are at the same voltage.
ii) have the same charge.
)
E
R1
E
+
R3
+
R2
R4
C1
C2
C3
C4
Phys 104: College Physics
Exam 1
Spring 2010
SHORT ANSWER
1)4 There are four graphs shown below. On each graph circle whether the device is ohmic or
nonohmic (circle your choice below each graph) AND whether the resistance is increasing,
decreasing or staying constant as voltage increases.
I
A
Ohmic
I
V
Non-Ohmic
Decreasing
R is ____________
B
C
I
V
Ohmic
Ohmic
Non-Ohmic
Constant
R is ____________
I
V
Non-Ohmic
Increasing
R is ____________
D
Ohmic
V
Non-Ohmic
Decreasing
R is ____________
2)4 A capacitor is connected to a battery. What happens to the following when the distance
E0A
between the plates is doubled? (circle one choice in each part).
C
=
a) C, the capacitance (increases, decreases, stays the same).
d
b) Q, the charge on the capacitor (increases, decreases, stays the same).
Q = CV
c) V, the voltage on the capacitor (increases, decreases, stays the same).
d) E, the electric field in the capacitor (increases, decreases, stays the same). V = Ed
A
B
C
+4µC
-8µC
3.0 cm
6.0 cm
3)4 For charges A = +4.0 µC, B = -8.0 µC and C = +3.0 µC shown,
a) What are the magnitude and direction of the electric force on C due to A?
b) What are the magnitude and direction of the electric force on C due to B?
c) What are the magnitude and direction of the electric force on C due to A and B?
a)
FEA =
kqAqC
(8.99 × 10 )( 4 × 10 )(3 × 10 ) = 120 N to the right
=
kqAqC
8.99 × 10 )( −8 × 10 )(3 × 10 )
(
=
= 60 N to the right
2
rAC
9
−6
−6
( 0.03)
2
9
−6
−6
b)
FEB =
c)
FEboth = FEA + FEB = 120 + 60 = 180 N to the right
2
rAC
( 0.06)
2
Phys 104: College Physics
Exam 1
Spring 2010
PROBLEMS: CHOOSE 3 OF 4.
1) For the charges -8.00 µC at (0.80,0.30) and +5.00
µC at (0, 0) and points A at (0, 0.30) and B at (0.40,
0.80). Find
a) the potential energy.
b) the electric potential at point A.
c) the electric potential at point B.
d) the change in potential energy for a 2 μC
charge moved from point A to point B.
e) the work done by the electric force to
move a 2 μC charge from point A to point B.
B
0.8 m
Y(m)
0.64 m
0.89 m
0.3 m
A
-8 μC
0.85 m
5 μC
0.4 m
0.8 m
X (m)
a) The potential energy is that of the two charges with U → 0 as r → ∞
(
)(
)(
8.99 × 109 5.0 × 10−6 −8.0 × 10−6
kq5q8
U=
=
r58
( 0.85)
b) The potential at A is due to both charges
(
) = -0.423 J
) (
⎛ 5.0 × 10−6
−8.0 × 10−6
kq5A kq8A
9 ⎜
VA =
+
= 8.99 × 10
+
⎜ ( 0.3)
r5A
r8A
( 0 .8 )
⎝
(
)
c) The potential at B is due to both charges
(
) (
⎛ 5.0 × 10−6
−8.0 × 10−6
kq5B kq8B
9 ⎜
+
VB =
+
= 8.99 × 10
⎜ ( 0.89 )
r5B
r8B
( 0.64 )
⎝
(
)
d) The change in potential energy in moving q from A to B
(
)(
) ⎞⎟ = 59.9 × 10
⎟
⎠
) ⎞⎟ = -61.9 × 10
⎟
⎠
)
ΔUA→B = qVB - qVA = 2.0 × 10−6 -61.9 × 103 − 59.9 × 103 = -0.244 J
It’s negative because A is at higher potential
so q loses potential energy in the move
e) The work done by the field in moving q from A to B is
positive since the particle loses potential energy (it
moves the way the field would move it)
WE −field = − ΔUA→B = + 0.244 J
3
3
= 59.9 kV
= -61.9 kV
Phys 104: College Physics
Exam 1
Spring 2010
2) A particle of m = 2.0 x 10-17 kg and Q = +12 µC is moving in the +x direction at v = 5.0 x 107m/s
when it enters a uniform field, E = 6.00 x 104 N/C between two charged plates that are 5.0 cm
long.
a) Draw an arrow indicating the direction of the E-field between the plates.
b) Indicate (with + and -) the charges on the plates
c) What is the magnitude of the acceleration of the particle due to the E-field?
d) How long does the particle stay between the plates.
e) Find the vertical deflection, d, of the particle as it leaves the plates.
5.0 cm
a) Electric field is down in the picture
b) Top plate is positive.
d
c) Use Newton’s Second Law
F = ma ⇒
(
)(
12 × 10−6 6.0 × 104
qE
F
a=
=
=
m
m
2.0 × 10−17
(
)
) = 3.6 × 10
16
m
s2
d) Use kinematics
x = vox t ⇒
t=
x
0.05
=
= 1 × 10 −9 = 1 ns
7
v0x
5.0 × 10
e) Use kinematics
y=
(
)(
1
1
aoy t 2 =
3.6 × 1016 1 × 10−9
2
2
)
2
= 1.8 × 10 −2 = 1.8 cm
Phys 104: College Physics
Exam 1
Spring 2010
3) For the circuit shown, where E = 20.0 V, R1 = R2 = 12.0 Ω,
R3 = 20.0 Ω and R4 = 40.0 Ω
a) Find the equivalent resistance for the network of resistors.
b) Find the voltage across resistor R3
c) Find the current through resistor R3
d) Find the power dissipated across resistor R3.
R12 = 6 Ω
R + R4
1
1
1
20 + 40
6
=
+
= 3
=
=
R34
R3
R4
R3R4
(20 )( 40 ) 80
R34 = 13.3 Ω
Resolve the right side series:
R2
R3
R4
E
a) Resolve each parallel pair of resistors:
2 (12 )
R + R2
1
1
1
1
=
+
= 1
=
=
R12
R1
R2
R1R2
(12)(12) 6
R1
R12
E
R34
R1234 = R12 + R34 = 6 + 13.3 = 19.3 Ω
b) Voltage across R3 is that across the R34 pair.
⎛ E ⎞
⎛ 20 ⎞
V12 = IR12 = ⎜⎜
⎟⎟ R12 = ⎜
⎟ (13.3) = 13.7 V
19
3
R
.
⎝
⎠
⎝ 1234 ⎠
c) Current through R3
⎛ V ⎞ ⎛ 13.7 ⎞
I3 = ⎜⎜ 34 ⎟⎟ = ⎜
⎟ = 0.69 A
R
20
⎠
⎝ 3⎠ ⎝
d) Power dissipated across R3
P3 = I32R3 = ( 0.69 ) (29 ) 0 = 9.50 W
2
R1234
E
Phys 104: College Physics
Exam 1
4) A heart defibrillator passes 10.0 A through a patient’s torso for 5.00 ms
in an attempt to restore normal beating.
a) How much charge passed through the patient?
b) What power was applied if 500 J of energy was dissipated?
b) What voltage was applied?
c) What was the path’s resistance?
a) Use the definition of current
I=
q
⇒ q = IΔt = (10 ) 5.0 × 10−3 = 5.0 × 10−2 = 50.0 mC
Δt
(
)
b) Use the definition of power
P=
U
500
=
= 1 × 105 = 100 kW
−3
Δt
5 × 10
c) The voltage is found from the power dissipated
P = IV ⇒
P
1 × 105
V=
=
= 1 × 10 4 = 10 kV
I
10
d) Use Ohm’s law
V = IR ⇒
V
1 × 104
R=
=
= 1000 Ω
I
10
Spring 2010