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:

4B SP12 REFERENCE: POTENTIAL FROM A RING OF CHARGE; CONSERVATION OF ENERGY:
QUIZ 3 #34 ; #9, #12, 14, 16

(a) (6 points) What is the total charge on the ring ?

(b) (11 points) Suppose a proton is located at the center of the ring, where it is released from rest. What will be the speed of the proton when it reaches a distance 0.20 m above the plane of the ring on the central perpendicular? See fig. 2 . The mass of the proton is:
.
The charge is:
.

(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 ?


            Fig. 1                               Fig. 2
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)
Ki + U i = Kf + Uf

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

                                                 

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.
Find the current in each resistor at t = 9.5 X10-6s after the switch is closed.
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
t = 9.5 X10-6s  to find i. This will give you the total i in the circuit. You can find the current in each resistor by doing the following:
i in 2-ohm = iRp/2

i in 4-ohm = iRp/4