READ ALL EXAMPLES ! SOME PROBLEMS ARE JUST LIKE AN EXAMPLE.

Quiz 3 Ch. 23--4** (try 12), 5*, 57*, 16**, 30**, 14**, 18** 19*, 43*, 49* , 39*, 60**, 61* (try #62 on Geiger counters !) , 35* , 38**, 79*, 46**, 55*

* DISCUSSIONS PROVIDED either in class or on the web /** EVEN ANSWERS PROVIDED

DISCUSSIONS

4. See example 1. We know the potential V at a distance r from a single point charge is k*q/r,. When another point charge Q placed at that location, the potential energy created is U = Q*V = KQ*q/r. Note: Change in U = -W, where W is the work by the field required to bring Q from infinity to a distance r from charge q.
We want to compute the work W(a to b) done by the field when the distance between Q and q is changed from r_a to r_b. W(a to b) = -change in U.
Now, change in U = k*Q*q/r_b   -   k*Q*q/r_a. Thus: W(a to b) = -change in U =
k*Q*q/r_a   -   k*Q*q/r_b .    Since the second radius ( at b) is smaller,
the work by field W(a to b) < 0.
Note: In moving the charge from a to b, the work W_you you do is -W(a to b). Thus W_you> 0.   Note work you do = change in U.
(b) Use conservation of energy (Ch. 7); KE means kinetic energy, U means potential energy, point 1 is when the two charges are close together, and point 2 is when they have moved back to the original distance : KE1 + U1 = KE2 + U2.

Assume KE1 = 0 since they start from rest. Thus KE2 = U1 - U2 = where U1 is the potential energy when the protons are 3.00 x10^-15m apart and U2 is when they are 2.00x10^-10 m apart. Thus KE2 > 0.

Below is another way of seeing the problem:

Identify: The work required is the change in electrical potential energy. The protons gain speed after being released because their potential energy is converted into kinetic energy.

(a) Set Up: Using the potential energy of a pair of point charges relative to infinity, we have

Execute: Factoring out the e² and substituting numbers gives

(b) Set Up: The protons have equal momentum, and since they have equal masses, they will have equal speeds and hence equal kinetic energy.

Execute: Solving for v gives = 6.78 ´ 10⁶ m/s

Evaluate: The potential energy may seem small (compared to macroscopic energies), but it is enough to give each proton a speed of nearly 7 million m/s.

5. (a) KEi + Ui = KEf + Uf.
KEi = initial kinetic energy of charge 1, the only charge that's moving in the problem. Ui = k*q1*q2/r _i .
KEf = final KE of charge 1. Uf = k*q1*q2/r_f . Find KEf and the final speed of charge 1.

(b) Use the initial kinetic and potential energy given in the problem at separation distance 0.800 m. KEi + Ui = KEf + Uf. In this case KEf = 0 for charge 1. Find Uf and r_f where charge 1 comes to rest.

16. .
Use work by field = -change in U . Now, change in U = q*change in V
Change in V = -E*(6.00 m) . Thus, work by field = -change in U = -change in V = + E*(6.00 m). The work by field is positive as you would expect for motion of a positive charge in the direction of the electric field.
(a) KEi + Ui = KEf + Uf.
Thus change in U = KEi - KEf and thus work by field = -change in U = KEf - KEi, where the initial kinetic energy is zero and the final KE is given.
(b) E*(6.00 m) = -change in U.
Below is another way of seeing the problem:

Identify: The work-energy theorem says

Set Up: Point a is the starting and point b is the ending point. Since the field is uniform, The field is to the left so the force on the positive charge is to the left. The particle moves to the left so and the work is positive.

Execute: (a)

(b) Point a is at higher potential than point b.

(c) , so

Evaluate: A positive charge gains kinetic energy when it moves to lower potential;

30.
Lets talk about where V = 0 first:
(a) (Q)----------------(2Q)
Let the charge Q be to the left of charge 2Q an indicated in the above schematic where the separation distance is d. Since both charge are positive, do think there is a place where V = zero? I think not !
(b)   (-Q)----------------(2Q)
Let the charge -Q be to the left of charge 2Q an indicated in the above schematic where the separation distance is d.
Certainly there is a point between the two charge where V = 0. Use:
V = 0 = k(-Q)/d'   + k(2Q)/(d - d'). Solve for d', the distance from the negative charge. We assume d' is positive; if d' is negative there is no solution between the charges.

Now look to the left of -Q and to the right of 2Q:
Left of -Q:   Let d' be the distance from the negative charge.
V = 0 = k(-Q)/d' + k(2Q)/(d + d') , where d' is the distance from negative charge. Solve for d'. We assume d' is positive; if d' is negative there is no solution to the left of negative charge.
Right of 2Q: Let d' be the distance from the positive charge.
V = 0 = k(-Q)/(d' +d) + k(2Q)/d' . Solve for d'. We assume d' is positive; if d' is negative there is no solution to the right of positive charge.

Now let's talk about the electric field magnitude E and where it could be zero on the line drawn between the charges.
a) (Q)----------------(2Q)
Let the charge Q be to the left of charge 2Q an indicated in the above schematic where the separation distance is d. Since both charges are positive, do think there is a place to the right of left of Q or the right of 2Q where E = zero? I think not !
Now look at a point between the charges.
E = 0 = kQ/d'²    - k*(2Q)/(d - d')². , where d' is the distance from charge Q. Find d' by turning this relation into into a quadratic equation and finding d' . We assume d' is positive; choose positive root..
(b)   (-Q)----------------(2Q)
Since charges are opposite in sign are positive, do think there is a place between there charges where E = zero? I think not !
Now look to the left of -Q and to the right of 2Q:
Left of -Q:   Let d' be the distance from the negative charge.
E = 0 = k(-Q)/d'² + k(2Q)/(d + d')² , where d' is the distance from negative charge. Solve for d'. We assume d' is positive; if d' is negative there is no solution to the left of negative charge.
Right of 2Q: Let d' be the distance from the positive charge.
E = 0 = k(-Q)/(d' +d)² + k(2Q)/d' ². Solve for d'. We assume d' is positive; if d' is negative there is no solution to the right of positive charge.

(c) Are both E and V zero at the same places? Explain.

Below is another way of seeing the problem:

Identify: For a point charge, . The total potential at any point is the algebraic sum of the potentials of the two charges. For a point charge, . The net electric field is the vector sum of the electric fields of the two charges.

Set Up: produced by a point charge is directed away from the point charge if it is positive and toward the charge if it is negative.

Execute: (a) so V is zero nowhere except for infinitely far from the charges. The fields can cancel only between the charges, because only there are the fields of the two charges in opposite directions. Consider a point a distance x from Q and from 2Q, as shown in Figure 23.30a. . The other root, does not lie between the charges.

(b) V can be zero in 2 places, A and B, as shown in Figure 23.30b. Point A is a distance x from and from 2Q. B is a distance y from and from 2Q. . .

The two electric fields are in opposite directions to the left of or to the right of 2Q in Figure 23.30c. But for the magnitudes to be equal, the point must be closer to the charge with smaller magnitude of charge. This can be the case only in the region to the left of . gives and .

Evaluate: (d) E and V are not zero at the same places. is a vector and V is a scalar. E is proportional to and V is proportional to . is related to the force on a test charge and is related to the work done on a test charge when it moves from one point to another.

Figure 23.30

14. work by field = -change in U.

Assume identical charges q are at the upper right and lower left of the square. Let us move the negative charge -q_o from the upper left to the lower right corner.
At upper left: U = -2k* q_o*q /a.
Find the U at the lower right and prove the change in U = 0; thus W = -change in U = 0.

Below is another way of seeing the problem:

Set Up: produced by a point charge is directed away from the point charge if it is positive and toward the charge if it is negative.

(b) V can be zero in 2 places, A and B, as shown in Figure 23.30b. Point A is a distance x from and from 2Q. B is a distance y from and from 2Q. . .

Figure 23.30

18. Here is a schematic of the arrangement of the two stationary, positive point charges:
(Q1)----------------(Q2)

The two charges above are separated by distance d. The electron starts at the center and is more attracted to the charge on the right end than the charge on the left end since we assume the right charge Q2 = |+3.00 nC (nano -coulomb); the electron moves rightward.
KEi + Ui = KEf + Uf.
KEi = 0.
Ui = -k*e*Q1/d_i' - k*e*Q2/(d - d_i') , where d_i'= d/2. Note d = 0.500 m .
Uf = -k*e*Q1/d_f ' - k*e*Q2/(d - d_f '), where d_f ' = (4/5)*d.
Solve for KEf and the final speed of the electron.

Below is another way of seeing the problem:

Identify: Apply

Set Up: Let and At point a, . At point b, and . The electron has and . since the electron is released from rest.

Execute: . .

Setting gives

Evaluate: . The negatively charged electron gains kinetic energy when it moves to higher potential.

19. Use V = kq/r and solve for r in each case.

43. (a) See example 23. 8. There is no charge inside the shell just like a solid, spherical conductor.
(b) Check back later.

49. (b) and 61 ((b), are like two birds of the same feather. We derived 61 (b) in class; let me take the opportunity to set up the derivation of of 49 (b)---leaving certain steps for you to derive E using Gauss's Law as we discussed in class.   We will discuss Part (a) of both problems in greater detail in class-- there are two ways---integration or the the book's recommendation of superposition.

49 (b) : First of all you need to prove, using Gauss's Law,   the electric field magnitude E (for   r_a< r < r_b ) = kq/r² . Then you must evaluate the following integral: Vb - Va =     -integral of Edr between a and b. In other words, integrate
- kqdr/r² between a and b. Clearly you will obtain the negative of the formula    Vab shown in the problem on page 809.

35. (a) Use the resultant formula of # 61(b): Vab = 2*k*lambda*ln (b/a) where b = 2.50 cm and a = 1.00 cm. Set 2*k*lambda*ln (b/a) = 575 (V). Find lambda.
(b) Redo with b = 3.30 cm and a = 1.00 cm and the lambda from the previous part. Your result should be greater than 575 (V).
(c) Note that you are setting a = b = 3.50 cm, independent of the distance between the probes which have the same value of r. What is ln 1?

60. Below is another way of seeing the problem:

Identify: Apply and to the sphere. The electric force on the sphere is . The potential difference between the plates is .