Phys 110 Physics Paper

PHYS 110 - Physics Test 3
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PHYS 110

Grossmont College
FALL, 2015
Campus: Grossmont
Department of Physics
Test 3
(Time allowed: DUE 11:35 AM, MONDAY 12/14/2015)
NOTE:

1. Attempt all questions.
2. You may get help (not answers) from tutors and/or classmates.
3. Make sure you show your work, or give an explanation of how you came to your answer. If there is no work or explanation there will be NO POINTS.

1. A force with magnitude F is needed to separate two charges by a distance r. What will be
1
the new force F ′ , needed to keep the two charges separated by a distance 2 r?
2. A a particle with charge of Q = 3.2 × 10−6…show more content…
Figure 2 shows a circuit with a resistor R = 1.0 × 103 Ω, and a battery with potential difference of VB = 5.0V .
(a) Solve the circuit in Figure 2. That is, find all unknown currents, voltages, and resistances.
(b) What is the direction of the current? Note: I want the direction of the conventional current. 5. Consider the circuit in Figure 3, where R1 = 5.00×102Ω, R2 = 1.00×103Ω, and VB = 10.0V .
(a) What is the equivalent resistance Req of the circuit?
(b) Solve the circuit. That is find all unknown currents, voltages, and resistances.
6. Consider the circuit in Figure 4, where R1 = 5.00×102Ω, R2 = 1.00×103Ω, and VB = 10.0V .
(a) Find the equivalent resistance Req of the circuit.
(b) Draw the simplified equivalent circuit.
(c) Solve the circuit. That is, find all unknown currents, resistances, and voltages.
7. You want to build a circuit which causes a ligh tbulb to turn on when you throw a switch.
So, you build the circuit if Figure 5. When the switch is connected to point A the capacitor charges, and when you connect the switch to point B the light goes on. Let the resistance of the light bulb be Rℓ = 1.50 ×103 Ω, the potential difference across the battery is VB = 10.0V , and the capacitor has a capacitance of C = 1.35 × 10−4 F…show more content…
The distance between the forces is given by the Coulombs law through the use of the formula F=kq1q2/r2.0.1newtons = 8.99*109*3.2*10-6*7.7*10-7/r2
R= 555.78
Answer to question 3
• Potential difference between the two plates is equal to velocity which is equal to 6.0*106m/s
• Force = mass *acceleration = 1.4*10-13*6.0*106 = -8254 nektons The speed of the particles are computed by the formula V=ED. This is equal to 8.5*10-6*0.15. This is equal to 84.1
Answer to question 4
Voltage = current *resistance. This implies that in this case while V is 5.0 and resistance is 1.0*103, current will be equal to 5/1.0*103, = 500 amps
B the direction of the conventional current provides the electric charge movement from the positive side of the battery to its negative side as in indicated in the diagram below

Answer to question 5
• This section focuses on the equivalent resistance of a circuit. The equivalent resistance will be equal to (5.0*102+1.00*103)2. = 306.5.
• With voltage of 10, the currrent will be equal to voltage/resistance = 10/306.5=