This test is 90 minutes long and has a total of 100 points.
[15 points] 1. Multiple choice questions. Each question is worth 3 points. Partial credit may be given for an incorrect answer if work is shown.
(i). A positively-charged insulating rod is brought close to, but not touching, a metal sphere that is initially uncharged. The sphere is grounded briefly, and then the ground connection is removed. The charged rod is then removed. The charge on the sphere is now:
[ ] Positive
[ ] Zero
[ x ] Negative
Electrons from the ground are attracted to the positive rod so they flow on to the sphere, giving it a net negative charge. Once the ground connection is removed they're stuck there, even after the charged rod is taken away.
(ii). Two identical charges are separated by a distance of 0.5 m. The force that the first charge exerts on the second is 360 N. The magnitude of each charge is :
[ ] 1.4 x 10-4 C
[ x ] 1.0 x 10-4 C
[ ] 1.4 x 10-8 C
[ ] 1.0 x 10-8 C
This is a straight-forward Coulomb's law problem. You just have to be careful to keep track of what's squared.
(iii). Two parallel wires carry currents in opposite directions. The force exerted by wire 1 on wire 2 is:
Below wire 1, the current running through wire 1 sets up a magnetic field pointing into the page (that's the second right hand rule). The first right hand rule can then be used to show that the force exerted on wire 2 by this magnetic field is down, which is away from wire 1.
(iv). If two identical 40 W light bulbs are connected in parallel to a wall socket and left on for a year, the cost would be about $70. If these same lamps are connected in series instead and left on for a year, how much would that cost?
Consider the power equation P = VI. When the bulbs are in parallel, the equivalent resistance is R/2, so the total current coming from the battery is I = 2 (V/R). In series the total resistance is 2R, so the total current is I = 0.5 (V/R). The current is reduced by a factor of 4 when the bulbs are in series, so the cost must also be down by a factor of 4. The cost for a year is therefore $17.50.
(v). Which of the following diagrams is NOT a possible representation of the electric field lines near a metal sphere?
Only the third case is not allowed. Field lines may not cross. Also, at equilibrium the field lines will be perpendicular to the surface of the conductor.
[5 points] 2. Moving a charge.
[3 points] (a) How much work is done in bringing the charge q from very far away to the point P if q is moved along the dashed line at constant speed?
Zero. The equation for potential is V = kQ/r. When r is very large (i.e., when q is very far away), the potential is small and there is zero potential energy associated with the charge q interacting with +Q and -Q. The potential at point P is also zero (the potentials from the + and - charges cancel) so placing q there again results in no potential energy. Neither the potential energy nor the kinetic energy changes, so no work is done.
[2 points] (b) How much work is done to bring the charge from very far away to point P by a more complicated path? Again the charge is moved at constant speed.
The work done is independent of the path. In other words, it doesn't matter how you get there; all the matters is where you started and where you end up. Same as part (a).
[20 points] 3. Model of a hydrogen atom.
One model of a hydrogen atom, which has one proton and one electron, suggests that the proton is at rest, and the electron orbits the proton in a circular orbit of radius r = 5.29 x 10-11 m. The gravitational interaction between the proton and the electron can be ignored.
[8 points] (a) What is the force exerted on the electron by the proton? State the magnitude, and draw an arrow on the diagram to indicate the direction of the force.
The proton exerts an attractive force on the electron, so the arrow should point from the electron toward the proton. To get the magnitude, just use Coulomb's law:
F = kqQ / r2
= (8.99 x 109)(1.60 x 10-19)2 / (5.29 x 10-11)2
= 8.2 x 10-8 N
[6 points] (b) What is the magnitude of the acceleration of the electron?
There is only one force acting on the electron, so just use F = ma:
a = F / m = 8.2 x 10-8 / 9.11 x 10-31
= 9.0 x 1022 m/s2
[6 points] (c) How long does it take the electron to complete one orbit?
The electron goes at constant speed in its circular orbit, so t = distance/speed:
the speed can be found using a = v2 / r, so
v = 2.18 x 106 m/s
This gives:
t = circumference / speed
= 3.32 x 10-10 / 2.18 x 106
= 1.5 x 10-16 s
[20 points] 4. Two conducting spheres.
Two conducting hollow spheres share a common center.
The inner sphere has a charge q = +1.5 x10-6 C and a radius of 10 cm.
The outer sphere has a charge Q = -4.0 x10-6 C and a radius of 30 cm.
[6 points] (a) On the diagram, sketch the electric field produced by the charged spheres. Be sure to indicate the direction clearly.
[5 points] (b) What is the electric field at a point 5 cm from the center of the spheres?
This point is inside both spheres, so it's unaffected by the charge on the spheres. The electric field is zero.
[4 points] (c) With a dashed line on the diagram, draw the equipotential line that passes through point P, which is a distance of 40 cm from the center of the spheres.
[ 5 points] (d) What is the electric potential at point P?
The potential is the sum of the potential at P due to the inner sphere plus the potential due to the outer sphere:
V = kq / r + kQ / r = k (q + Q) / r
Potential is NOT a vector, so keep all the signs.
V = 8.99 x 109 (-2.5 x 10-6) / (0.4)
= -56200 volts
[20 points] 5. A multi-loop circuit
[4 points] (a) Apply Kirchoff's loop rule to loop 1. Write down the equation; from your equation, solve for I1. Note that I3 = 0.500 A.
Starting in the bottom left corner and going around the loop clockwise gives:
+5 -5I1 - 2I3 -3 = 0
Solving for I1 gives:
1 = 5I1 so I1 = 0.2 A
[4 points] (b) Apply Kirchoff's loop rule to loop 2. Write down the equation and solve for I2.
Starting in the bottom right corner and going around the loop counter-clockwise gives:
+7 +10I2 - 2I3 -3 = 0
Solving for I2 gives:
3= -10I2 so I2 = -0.3 A
[4 points] (c) Apply Kirchoff's junction rule at one of the junctions. (Again, write down the equation.) You can use this equation to check whether your answers for parts (a) and (b) are correct.
It doesn't matter which junction, A or B, you use:
I1 = I2 + I3
Subbing in the values found above gives:
0.2 = -0.3 + 0.5
Looks good.
[4 points] (d) What is the magnitude of the potential difference between points A and B in the circuit?
There are three ways to get from B to A; they should all give you the same answer. If you go up through the middle branch, the voltage increases by 3 V when you cross the battery. The voltage across the resistor is 1 V (applying V = IR), and the top end is at higher potential because the current is flowing down. 3V + 1V = 4V, so A is 4 V higher than B. The answer is 4 V.
[4 points] (e) How much power is the 5 V battery putting into the circuit?
For a battery the simplest thing to do is to apply P = VI. For this battery the current is 0.2 A, so the power is P = 5 (0.2) = 1 V.
6. [ 20 points ] A cyclotron.
The first cyclotron was built in 1932; it's a compact device for accelerating charged particles to very high speeds.
A sample cyclotron is shown above. The charges (protons, in this case) are accelerated by applying a potential difference V across the gap. The protons pass through the gap several times, being accelerated each time. The potential difference must be reversed each time the protons cross the gap; otherwise, the speed gained going from left to right would be lost going from right to left.
To swing the protons back into the gap, a uniform magnetic field is present in each of the D-shaped regions, known as dees. The magnetic field is the same in each dee, and is perpendicular to the velocity of the protons.
[5 points] (a) The protons start from rest at point a. If they reach point b with a speed of 3.2 x 105 m/s, what is the potential difference V across the gap? (The gap can be treated as a parallel-plate capacitor.)
Applying conservation of energy gives:
qV = 0.5 mv2
q = 1.6 x 10-19 C; v = 3.2 x 105 m/s; and m = 1.67 x 10-27 kg.
Solving for V gives V = 534 volts.
[4 points] (b) The width of the gap (the distance between the dees) is 4.4 x 10-3 m . What is the magnitude of the electric field in the gap?
Treating the gap as a parallel-plate capacitor means the equation E = V/d applies. This gives:
E = 534 / 4.4 x 10-3 = 1.21 x 105 V/m
Although the sign of the potential difference across the gap must be changed each time the protons cross it, the direction of the magnetic field in the dees can be left fixed.
[3 points] (c) The protons follow a counter-clockwise path, covering half a circle, in the dees. What direction is the magnetic field in the dees?
[ ] left
[ ] right
[ ] up
[ ] down
[ x ] into the page
[ ] out of the page
Apply the right hand rule. This will work at any point, but consider it at point b. The velocity is to the right, and the force is up. Holding your hand flat with the thumb pointing right (v) and the palm up (F), the fingers go into the page (B).
[4 points] (d) The straight-line distance from b to c is 1.20 cm. If the protons reach point b with v = 3.2 x 105 m/s, what is the magnitude of the magnetic field B in the dees?
Apply (or derive!) the equation for the radius of a charged particle moving in a magnetic field:
r = mv / (qB). Note that the 1.20 cm is the diameter, so the radius is 0.6 cm, which is 0.006 m. Solving for B gives:
B = mv / (qr) = 1.67 x 10-27 (3.2 x 105) / [1.6 x 10-19)(0.006)]
= 0.557 T
[4 points] (e) The time for protons to travel through half a circle in a dee does not depend on the velocity. This is convenient, because the voltage across the gap can then be reversed in the same amount of time each time. It's easy to show that the time for protons to travel half a circle through a dee is:
Calculate the time t for protons in this particular cyclotron.
Just plug in the numbers, nothing complicated about it. Doing that gives:
t = 5.89 x 10-8 s
This is a short time, but that's enough time to reverse the voltage.