Second Hour Exam
for
PHY 1361
March 8, 2002

Equation sheet

F = k Qq/r2
E = F/q
E = k Q/r2
V = U/q
V = k Q/r
k = 9 x 109 N-m2/C2
k = 1/4

=
Q/
E = k Q/r2
E = k [Q/a3] r
E = 2 k /r
E = /2
E = V/s
C = Q/V
Ceq = C1 + C2 + C3
1/Ceq = 1/C1 + 1/C2 + 1/C3
U = (1/2) Q V = (1/2) C V2 = (1/2) Q2/C

Second Hour Exam
for
PHY 1361
March 8, 2002

Name:

1. __________
2. __________
3. __________
4. __________
CQ ________
Total __________

Again, pace yourself . Don’t spend too much time on one problem. Write what you know and go on to another problem. If you have time, you can come back and think more about a problem. I can’t give partial credit for a blank page.

Calculus is important. I chose to use the parallels with problems 1 and 2 describing electric field and electric potential. This choice left insufficient time for a good calculus problem. Such a calculation from an extended charge distribution will surely appear on the final.

Give a sigh of relief and enjoy your Spring Break!

1. Charges q1, q2, and q3 have the values and locations shown here. What is the electric field (magnitude and direction) at point P as shown here?

The net electric field is given by

E = E1 + E2 + E3

Be sure that you remember that this is a vector equation!

Now we use Coulomb’s Law to find the individual electric fields E1, E2, and E3 due to the individual charges q1, q2, and q3.

E1 = k q1/r12
E1 = (9 x 109 N m2/C2) (1 x 10 - 6 C) / (0.40 m)2
E1 = 5.63 x 104 N/C

This is the magnitude of the electric field. From the diagram, we see that it points to the left. We can write this specifically as

E1x = - 5.63 x 104 N/C and E1y = 0

And we continue by calculating E2, the electric field due to charge q2,

E2 = k q2/r22
E2 = (9 x 109 N m2/C2) (2 x 10 - 6 C) / (0.50 m)2

We found the distance d = 0.50 m from charge q2 to point P by using the Pythagorean Theorem.

E2 = 7.20 x 104 N/C

Notice the direction from the diagram. E2 = 7.20 x 104 N/C is the magnitude of E2 but we need the components to specify the direction or to continue with our vector addition problem.

E2x = E2 cos 37° = (0.80) E2
E2x = (0.80)(7.20 x 104 N/C)
E2x = 5.76 x 104 N/C
E2y = E2 sin 37° = (0.60) E2
E2y = (0.60)(7.20 x 104 N/C)
E2y = 4.32 x 104 N/C

E3 = k q3/r32
E3 = (9 x 109 N m2/C2) (3 x 10 - 6 C) / (0.30 m)2
E3 = 3.00 x 105 N/C
E3 = 30.0 x 104 N/C

As before, this is the magnitude and we need to take care of the vector nature of this by writing

E3x = 0 and E3y = 30.0 x 104 N/C

Now we add (as vectors!) the three electric fields due to the three charges;

E = E1 + E2 + E3

But this is true only as a vector equation and really means

Ex = E1x + E2x + E3x and Ey = E1y + E2y + E3y
Ex = (5.63 +5.76+0)x104 N/C and Ey = (0+4.32+30.0)x104 N/C
Ex = 11.39 x104 N/C and Ey = 34.32x104 N/C

Now that we have the components, we must go back and reconstruct the net electric field E, finding the magnitude E and the direction .

tan = opp/adj =34.32/11.39 = 3.01
= 79.6°
E = SQRT[Ex2 + Ey2]
E = SQRT[(11.39)2 + (34.32)2] x 104 N/C
E = 36.2 x 104 N/C

Again, I have used SQRT[...] to mean square root of [...] because that is easier to type.

2. Charges q1, q2, and q3 have the values and locations shown here. What is the electric potential at point P as shown here?

Of course, this is the same arrangement of charges that we had in the previous problem. Now, we want to find the electric potential. Remember, electric potential is a scalar.
The electric potential due to a single point charge q is

V = k q/r

So we begin by finding the electric potential a point P due to each one of the charges in turn and then add up all these scalar values for the total,

Vtot = V1 + V2 + V3

V1 = k q1/r1
V1 = (9 x 109 N m2/C2) ( - 1 x 10 - 6 C) / (0.40 m)
V1 = - 2.25 x 104 J/C

V2 = k q2/r2
V2 = (9 x 109 N m2/C2) ( 2 x 10 - 6 C) / (0.50 m)
V2 = 3.60 x 104 J/C

V3 = k q3/r3
V3 = (9 x 109 N m2/C2) ( 3 x 10 - 6 C) / (0.30 m)
V3 = 9.00 x 104 J/C

Remember, electric potential is a scalar so these electric potentials add as ordinary numbers,

Vtot = V1 + V2 + V3
Vtot = ( - 2.25 + 3.60 + 9.00 ) x
104 J/C
Vtot = 10.4 x 104 J/C
Vtot = 10.4 x 104 V

3. (24.58) Two infinite, nonconduciting sheets of charge are parallel to each other as sketched here. The sheet on the left has a uniform surface charge density + ( + sigma ) and the one on the right has a uniform charge density - ( - sigma).
Show all your work and calculate the value of the electric field at points
(a) to the left of,
(b) in between, and
(c) to the right of the two sheets.

4. The capacitor circuit shown here is connected to a 12-volt power supply.
(a) Calculate the equivalent capacitance of the circuit shown here.
(b) What charge will flow out of the battery?
(c) What charge is stored on the 2.0 F capacitor?
(d) What voltage is across the2.0 F capacitor ?

We begin by replacing the 2.0 F and 6.0 F piece with a single equivalent capacitor of value Ceq(1).

1/Ceq(1) = 1/(2.0 F) + 1/(6.0 F)
1/Ceq(1) = [1/2.0 + 1/6.0]/F
1/Ceq(1) = [0.5 + 0.1667]/F
1/Ceq(1) = 0.6667/F
Ceq(1) = (1/0.6667)F
Ceq(1) = 1.5 F

Now, this Ceq(1) = 1.5 F capacitor is in parallel with the 3.0 F capacitor so the single equivalent capacitor that can replace both of these is

Ceq(2) = 1.5 F + 3.0 F
Ceq(2) = 4.5 F

When this equivalent capacitor is connected to a 12-volt power supply, the charge that flows out of the battery or power supply onto this capacitor is

C = Q/V
Q = C V
Qtot = (4.5 x 10 - 6 F)(12 V)
Qtot = 5.4 x 10 - 5 C

Where does this charge reall go? Some of it goes onto the 3.0 F capacitor and some goes onto the earlier equivalent capacitor Ceq(1). The sum of these two charges is Qtot. We have enough information to calculate the charge on the 3.0 F capacitor so we begin with that. This 3.0 F capacitor has the full 12 volts across it.

Q3 = C3 V3
Q3 = (3.0 x 10 - 6 F)(12 V)
Q3 = 3.6 x 10 - 5 C
Qtot = Q2 + Q3
Q2 = (5.4 - 3.6) x 10 - 5 C
Q2 = 1.8 x 10 - 5 C

Q2 is the charge that flows onto the first equivalent capacitor Ceq(1) and it is also the charge on the 2.0 F capacitor (and the same charge is on the 6.0 F capacitor altho’ we do not need that information).
Now for the voltage across this 2.0 F capacitor.

C = Q/V
V = Q/C
V2 = Q2/C2
V2 = (1.8 x 10 - 5 C)/(2.0 x 10 - 6 F)
V2 = 9.0 V
V2 = 9.0 volts

And that will mean that

V3 = 3.0 volts

altho’ we do not need that information either.

Conceptual Questions:

1.
A ballloon is negatively charged by rubbing and then clings to a wall. Does this mean that the wall is positively charged?

The balloon is attracted to and held in place by charge polarization or charge induction -- even tho' the wall is electrically neutral. We talked about this -- and demonstrated it -- in class.

2. Two solid spheres, both of radius R, carry identical total chares, Q. One sphere is a good conductor while the other is an insulator. If the charge on the insulating sphere is uniformly distributed throughout its interior volume, how do the electric fields outside these two spheres compare? Are the fields identical inside the two spheres?

Outside the radius R, the electric fields are identical.

Inside the radius R, the electric field inside the conductor is zero. Inside the radius R, the electric field inside the conductor increases linearly with radius from zero at the center.

3. Give a physical explanation of the fact that the potential energ of a pair of like charges is positive whereas the potential energy of a pair of unlike charges is negative.

Think back to the case of gravitational potential energy ( U = m g h ). Suppose that you choose or define the "reference point" for zero gravitational potential energy ( U = 0 ) to be on top of a table. If you have a book above this point it has positive gravitational potential energy and will fall back to U = 0 if you release it. If you have a book below this point it has negative gravitational potential energy and you must do work on it to get it back to U = 0.

Now, look at the electrical equivalent. We have choosen or defined U = 0 to be the state where the two particles are far, far apart. If we have two like charges that are near each other, they will go to this state if we release them. That is, they will repel each other and push each other far, far away. That means they had positive potential energy when they were nearby. That is not the case for two unlike charges. If we find them nearby, we must do work on them to get them to the state we have defined as U = 0. That means they had negative potential energy when they were nearby.

4. The plates of a capacitor are connected to a battery. What happens to the charge on the plates if the connecting wires are removed from the battery? What happens to the charge if the wires are removed from the battery and connected to each other?

If the wires to the battery are disconnected, the charge remains on the plates -- and the voltage across the plates remains the same. If the wires are connected to each other, current will flow and the capacitor will discharge. Then there will be no voltage across the capacitor nor any charge on the plates.