Caution: These are not the only kinds of problems that may occur on the final exam. See the weekly quizzes for a greater variety of possibilities.
PHYSICS 7
MIDTERM EXAM (Ch. 15 – 21)
1. A +300 C charge is placed on the zero mark of a meter stick, while a –100 C charge is placed at the 30-cm mark. Calculate the net force on a –25 C charge placed at the 100-cm mark. (One C is one microCoulomb = 1X 10 –6 C) (k = 8.99 X 109 N-m2/C2)
Answer: 21.5 Newton to the left
2. Two parallel conducting plates are 1.50 cm apart and have a uniform electric field between them with a magnitude of 1500 N/C. (a) How much work is done in moving a proton from the negative plate to the positive plate? (b) What is the potential difference between the plates? (Charge on one proton , q = 1.60 X 10-19 Coulomb)
Ans (a) 3.60 X 10-18 Joules
3.(3 pts) If you are given three different capacitors, C1, C2, and C3, how many different combinations of capacitance can you produce, using every capacitor in each circuit?
Ans: 8
4. A parallel-plate capacitor filled with a paper dielectric has plates of area 2.30 X 10 -4 m2 separated by 1.50 mm. The capacitor is connected to a 12.0-V battery. (a) What is the charge on the capacitor? (b) What is the magnitude of the uniform electric field between the plates?
( for paper dielectric, = 3.7) (permittivity of free space, o = 8.85 X 10 –12 C2/N-m2)
Ans: (a) 6.03 X 10-11 C or 60.3 picoCoulombs (b) 8000 N/C
5.(3 pts) Suppose a long wire runs along the top of the marker board in this classroom, carrying current from left to right (roughly west to east). The current will create a magnetic field everywhere in the room. What will be the direction of the magnetic field at the back wall of the room? (The answer will be north, south, east, west, up or down.) Ans (a) Down
NAME______
6.Seven 120- resistors are connected to a 16-V battery as shown below. Find the voltage drop across the indicated resistor.
Ans: 4.00 volts
7. Write the Kirchhoff’s rules equations for the circuit shown below and use them to find the current through the 8.0- resistor.
Ans: Equations and 2.40 Amps
8.A strong wind blows down a high-voltage wire which hangs down a distance of 3.0 meters, vertically. The lower end of the wire touches a puddle of water, which draws a large current of 12.0 amperes. The Earth’s magnetic field at that location points North, but at a 22-degree angle above the horizontal. The magnitude of the Earth’s magnetic field is 1.50 X 10-4 Tesla. Calculate the net force on the wire due to the magnetic field.
Ans: 5.01 X 10-3 N
9.In the circuit RC shown below, the capacitor is initially uncharged. Find the charge on the capacitor 3.00 seconds after the switch is closed.
Ans: 287 F
10. A copper wire, connected in series with a 100- resistor, is formed into a circular loop whose radius is 8.29 cm. The loop of wire is lying horizontally on the ground. There is a vertical magnetic field in the area of strength B = 2.30 Teslas. (a) Calculate the magnetic flux through the coil. (b) If the magnetic field is suddenly reversed, so that in 0.50 second it points in the opposite direction, find the current in the resistor during this time. (c) what will be the direction of the current, through the resistor, left or right?
Ans: (a) 0.497 wb (b) 1.99 mA (c) Right
11. (21 points) A 17.7-F capacitor, a 0.60-Henry inductor, and a 100- resistor are all connected in series with a 120-volt (rms), 60-Hz power supply. Calculate:
a) the capacitive reactance of the capacitor Ans: 150
b) the inductive reactance of the inductorAns: 226
c) the total impedance of the circuit;Ans: 126
d) the current through the circuit;Ans: 0.955 A
e) the phase angle between the voltage and the current in the power supply;
Ans: 37.2
f) the power used by the circuit;Ans: 91.2 W
g) the voltage drop across the inductor.Ans: 216 V
1 F = 1 X 10-6 F
Name______
Work = force x distance, potential diff, V = work/charge
Newton’s 2nd Law: F= ma
Pythagorean Theorem: a2 + b2 = c2
Coulomb’s Law:
Electric field, E =
Point Charge:
Electric Flux,
Work done by electric field: W = Fx
Potential near a point charge: V =
E = PE = C =
Parallel plate capacitor:C = oA/d
Capacitors in parallel:C = C1 + C2 + C3 + ......
Capacitors in series:
1/C = 1/C1 + 1/C2 + 1/C3 ....
ke = 8.99 X 109 N-m2/C2
o = 8.8r X 10-12 C2/N-m2
Capacitance, C Q/V
with dielectric, C Q/V
Parallel plate capacitor:
C = koA/d
Energy, E = ½ CV2 = ½ QV = ½ Q2/C
Capacitors in parallel: Ceq = C1 + C2 + C3 + ....
Capacitors in series:
Definition of resistance: R = V/I
Resistance of a resistor: R = l/A where is resistivity lis length and A is cross-sectional area
Current = charge/time: I(amps) = Q/t
Power = IV = I2R = V2/R
Variation of resistance with temperature:
R = RoT where is temperature coefficient
R = R0(1 + T)
Ohm’s Law:
V = IR
Internal Resistance
terminal voltage, V = - Ir
Series Resistors
Requivalent = R1 + R2 + R3 + ....
Parallel Resistors
Q = CV
RC Circuits:
F = qvBsin()
F = B I lsin() (l is length of wire)
(torque) = BIAsin()
r =
Circumference of a circle: C = 2r
Loop:
Solenoid:
Irms = Imax/ , vrms = v max/
Xc =
XL = 2fL
______
Z = R2 + (XL – XC)2
Vtotal = IZ
P = I cos()
P = I2R = I2Z cos()
cos() = R/Z
tan() = (XL – XC)/R
fresonant=
1