EXAM 1 – 100 points



EXAM 1 IUPUI Physics 219

Dr. Edward Rhoads Summer 2007

NAME__________________________________________________

Part I – Multiple Choice Conceptual Questions

Directions: Circle for the one alternative that best completes the statement or answers the question. Unless otherwise stated, assume ideal conditions (no air resistance, uniform gravity, etc.) 2 points each

1) Two charged objects attract each other with an electric force F. Suppose that the distance between them is doubled. The new electric force equals

(A) 2F (B) ½F (C) 4F

(D) 1/8F (E) 1/4 F

2) A higher Capacitance tells you that:

(A) you collecting charge more efficiently

(B) you can collect less charge

(C) you are collecting charge less efficiently

(D) you are using a dielectric with a lower dielectric constant

(E) none of the above

3) Electric field lines ALWAYS

(A) point toward decreasing values of potential.

(B) point in the same direction as the electric force.

(C) are parallel to equipotential surfaces.

(D) start on negative charges and end on positive charges.

(E) point directly to a charge.

4) A proton, released at rest in an electric field (with an assortment of charges), ALWAYS moves:

(A) along an equipotential surface. (C) in a straight line directly to the nearest electron

(B) nowhere. (D) down an electric field line (high to low potential).

(E) up an electric field line (low to high potential).

5) Electric Potential is in units of:

(A) Volts (B) Joules (C) Newton / Coulomb

(D) Coulombs (E) Watts

6) A resistance is a measure of a devices:

(A) ability to store charge (B) ability to limit flow of charge

(C) ability to resist heating (D) ability to produce force

(E) ability to resist corrosion

7) In general when two resistors of unequal resistance are connected in series,

(A) the heat generated by each resistor is the same.

(B) the current through each resistor is the same.

(C) the voltage across each resistor is the same.

(D) the equivalent resistance is smaller than the bigger resistance.

(E) more current flows out of the resistors than flows into the resistors

8) When current passes through a resistor:

(A) the current decreases

(B) the potential decreases

(C) the potential increases

(D) the voltage increases

(E) the current increases

9) Electric companies bill their customers for

(A) current (B) voltage (C) energy

(D) power (E) force

10) In general, when two resistors of unequal resistance are connected in parallel,

(A) the heat generated by each resistor is the same.

(B) the current through each resistor is the same.

(C) the voltage across each resistor is the same.

(D) the equivalent resistance is larger than the bigger resistance.

(E) more current flows out of the resistors than flows into the resistors

Part II – Word Problems – 20 points each

Directions: Solve each problem using the formulas listed on the last page. Show your work to receive full credit. Box your final answer, clearly label which question it goes with, and record its appropriate unit.

1) y

+4.0 nC

6.0 m

+5.0 nC

x

0 8.0 m

–6.0 m

–4.0 nC

For the charge configuration shown above, compute

(A) find the x and y components of the vector of the electric force on the -4.0-nC charge due to the +4.0-nC charge.

(B) find the x and y components of the vector of the electric force on the -4.0-nC charge due to the +5.0-nC charge.

(C) find the magnitude of the net electric force on the -4.0-nC charge, and

(D) draw the direction of the net electric force acting on the -4.0-nC charge on the diagram above (draw the arrow starting from the charge please).

2) In class we examined what it would take to trap lightning in a capacitor. Suppose you want to use a capacitor to run a house porch light overnight. However, you are space limited so the biggest area you want to use is about 6 square meters. Cost is an issue, and dielectrics help. For the cost, water is the best dielectric with a dielectric constant of 80. Since this system uses house power it should be set to the same voltage as your wall sockets which is 120 Volts.

Calculate:

(A) the energy used by the 10 watt fluorescent light bulb (in terms of Joules) over a 6 hour period.

(B) the charge stored to provide this energy

(C) the capacitance of the capacitor

(D) the separation distance between the plates

(E) is your answer for D a plate separation distance that could actually be reliably manufactured? Why or why not?

3) 6.0 Ω 8.0 Ω

A 5.0 Ω B

4.0 Ω 3.0 Ω

A) Calculate the total equivalent resistance between points A and B of the resistor network shown below.

B) If the current going through the 6.0 Ω resistor is 16 A then what is the current running through the 5.0 Ω resistor?

4) EVIL SCIENTIST CHALLENGE QUESTION:

One day you take a wrong turn coming to class and end up in the office of an evil scientist. He has hatched a scheme to take control of the internet! In the diagram below the 5 Ohm resistor is in fact the device he will use to take control of the internet. You notice that if the current through the device is just barely more than 10 amps then the device will burn out (the box the device is in reads “Warning, do not exceed 10 Amps). You must stop this device it is too late! However, if you get the current to be a lot more than 10 Amps it will explode, killing you. Only you can stop this evil genius. To short out this device you want to put in a voltage device next to the 5 Ohm resistor. This test will self destruct in 5 minutes.

Use Kirchhoff’s Rules to compute the currents I1, I2, and the needed value of V in the circuit shown. Hint: how large should I3 be?

I1 30 V

0.0 Ω 8.0 Ω

1.0 Ω add V

I2 here ↓

(evil device!)

0.0 Ω 5.0 Ω

> 10 A

60 V

EXAM 1 Formulas and Constants

[pic] [pic] [pic] [pic] [pic]

Fnet = ma weight = mg g = 9.80 m/s2 K = ½mv2 W = FΔx

Fe = -k q1 q2 / r2 Fe = qE E = -k q / r2 k = 8.99×109 N-m2/C2

U = qΔV = -k q1 q2 / r K = –qΔV [pic] Q = CΔV

C = κ A ε0 / d Cp = C1 + C2 + C3 [pic]

E = Q / (κ ε0 A) U = 1/2 QV = ½C(ΔV)2 = ½ κ A ε0 d E2 τ = RC

ε0 = 8.85×10–12 F/m melectron = 9.11×10–31 kg mproton = 1.67×10–27 kg e = 1.60×10–19 C

[pic] [pic] V = IR VT = E ± Ir [pic]

Iin = Iout Rs = R1 + R2 + R3 [pic]

Ibranch = Iintonode * (Rallbranches / Rbranch)

SI prefixes: G = 109 M = 106 k = 103 c = 10–2

m = 10–3 μ = 10–6 n = 10–9 p = 10–12

If 1/X = 1/Y + 1/Z then X = YZ / (Y + Z)

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