From Spring 1998 |
Answers to Real T3 |
1. (20 points) A ring of radius R =
0.20 m carries a uniformly distributed positive charge, as shown below in fig. 1. The
linear charge density of the ring is: (a) (6 points) What is the total charge on the ring ? (c) ( 3 points) What will be the speed of the proton in part (b) when it reaches an infinite
distance above the plane of the ring on the central perpendicular ? |
Solution Outline: (a) (b) Use conservation of energy from Ch. 8. That is the simplest method ! Ki + U i = Kf + Uf 0 + qVi = Kf + qVf 0 + (+e)Vi = Kf + (+e)Vf 0 + (+e)kQ/( R2)½ = Kf + (+e)kQ/( 0.202 + R2)½ (+e)kQ/R = ½mvf2 + (+e)kQ/ ( 0.202 + R2)½ Solve for vf . (c) 0 + qVi = Kf + qVf 0 + (+e)Vi = Kf + (+e)Vf 0 + (+e)kQ/( R2)½ = Kf + 0 (+e)kQ/R = ½mvf2 Solve for vf . |
2. (20 points) A spherically shaped capacitor
of inner radius a = 1.00 m and outer radius b = 2.00 m is used to heat up a
5.00 mg sample of metal. Note: 1 mg = 0.001g. (a) (3 points) Derive the formula for the capacitance of the spherical system in terms of the symbols a, b, and other relevant symbols. If you can't derive the formula, please go on to the next part and use the result for the capacitance hopefully written on your sheet of notes. (b) (17 points) What is the voltage difference V of this capacitor in
order to heat the metal from 293 K up to 600 K ? The metal has specific heat 130.0 J/kg0C
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Solution Outline: (a) See the derivation I did in lecture and in the text in example 26.3. You must derive such a formula in a test situation! (b) ½ CV2 = mC(600 - 293) . Solve for V. |
3. ( 20 points) What is the required resistance of an immersion heater that will increase the temperature of 1.6 kg of water from 50C to 550C in 10 min (=600 seconds) with a voltage V = 110 (V) ? The specific heat of water is 4190 J/(kg K.) |
Solution Outline: rate of heat generation = V2/R . heat absorption/time= mC(55 - 5)/600 . V2/R = mC(55 - 5)/600 . Solve for R. |
4. EXTRA CREDIT (10 points) The
switch is closed at t = 0. The battery voltage is: The two resistors are in parallel.
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Solution Outline: First transform the resistors into their parallel equivalent: 1/Rp = 1/2 + 1/4. Thus, Rp = 4/3 ohms. Second, use the formula for
the current in a charging circuit given by 28.15. Use Rp, C and the given battery voltage
in the formula. Evaluate the formula at i in 4-ohm = iRp/4
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