**wangsness electromagnetic fields solutions - Cap - 1**

wangsness electromagnetic fields solutions

(Parte **1** de 2)

1. Eq. 21-1 gives Coulomb’s Law, Fk q qr , which we solve for the distance:

k q q r

2. (a) With a understood to mean the magnitude of acceleration, Newton’s second and third laws lead to kg m s

m s c hc h

(b) The magnitude of the (only) force on particle 1 is q q q

Inserting the values for m1 and a1 (see part (a)) we obtain |q| = 7.1 u 10

3. The magnitude of the mutual force of attraction at r = 0.120 m is q q F k

4. The fact that the spheres are identical allows us to conclude that when two spheres are in contact, they share equal charge. Therefore, when a charged sphere (q) touches an uncharged one, they will (fairly quickly) each attain half that charge (q/2). We start with spheres 1 and 2 each having charge q and experiencing a mutual repulsive force

2/Fkqr . When the neutral sphere 3 touches sphere 1, sphere 1’s charge decreases to q/2. Then sphere 3 (now carrying charge q/2) is brought into contact with sphere 2, a total amount of q/2 + q becomes shared equally between them. Therefore, the charge of sphere 3 is 3q/4 in the final situation. The repulsive force between spheres 1 and 2 is finally q q q F F k k F

F q Q qr wherer is the distance between the charges. We want the value of q that maximizes the functionf(q) = q(Q – q). Setting the derivative /dFdqequal to zero leads to Q – 2q = 0,

5. The magnitude of the force of either of the charges on the other is given by

6. The unit Ampere is discussed in §21-4. Using i for current, the charge transferred is 462.510A2010s0.50C.qit u

7. We assume the spheres are far apart. Then the charge distribution on each of them is spherically symmetric and Coulomb’s law can be used. Let q1 and q2 be the original charges. We choose the coordinate system so the force on q2 is positive if it is repelled by q1. Then, the force on q2 is

F q q k q q

wherer = 0.500 m. The negative sign indicates that the spheres attract each other. After the wire is connected, the spheres, being identical, acquire the same charge. Since charge is conserved, the total charge is the same as it was originally. This means the charge on each sphere is (q1 + q2)/2. The force is now one of repulsion and is given by

F r k q q r b q q q q 1 d id i b g .

We solve the two force equations simultaneously for q1 and q2. The first gives the product

q q r Fk

2b g b g and the second gives the sum

q q r Fk where we have taken the positive root (which amounts to assuming q1 + q2t 0). Thus, the product result provides the relation q q u which we substitute into the sum result, producing

Multiplying by q1 and rearranging, we obtain a quadratic equation

The solutions are

u r u u |

If the positive sign is used, q1 = 3.0 u 10

–6 C, and if the negative sign is used,

(b) If we instead work with the q1 = –1.0 u 10 –6

Note that since the spheres are identical, the solutions are essentially the same: one sphere originally had charge –1.0 u 10 –6

What if we had not made the assumption, above, that q1 + q2t 0? If the signs of the charges were reversed (so q1 + q2 < 0), then the forces remain the same, so a charge of

C on one sphere and a charge of –3.0 u 10 –6 C on the other also satisfies the conditions of the problem.

q Qq q Q F

which (if we demand F2y = 0) leads to /1/22Qq . The result is inconsistent with that obtained in part (a). Thus, we are unable to construct an equilibrium configuration with this geometry, where the only forces present are given by Eq. 21-1.

8. For ease of presentation (of the computations below) we assume Q > 0 and q < 0 (although the final result does not depend on this particular choice).

(a) The x-component of the force experienced by q1=Q is

Q Q q Q Q q Q q F which (upon requiring F1x = 0) leads to /||22Qq , or /2.83.Qq (b) The y-component of the net force on q2 = q is

9. The force experienced by q3 is

| || | | || | | || |1 | j (cos45 i sin 45 j) i |

q q q q q q F F F F

(a) Therefore, the x-component of the resultant force on q3 is q q

(b) Similarly, the y-component of the net force on q3is q q

10. (a) The individual force magnitudes (acting on Q) are, by Eq. 21-1,

which leads to |q1| = 9.0 |q2|. Since Q is located between q1 and q2, we conclude q1 and q2 are like-sign. Consequently, q1/q2 = 9.0.

(b) Now we have

which yields |q1| = 25 |q2|. Now, Q is not located between q1 and q2, one of them must push and the other must pull. Thus, they are unlike-sign, so q1/q2= –25.

. q q q q F F F k k

We note that each term exhibits the proper sign (positive for rightward, negative for leftward) for all possible signs of the charges. For example, the first term (the force exerted on q3 by q1) is negative if they are unlike charges, indicating that q3 is being pulled toward q1, and it is positive if they are like charges (so q3 would be repelled from q1). Setting the net force equal to zero L23= L12and canceling k,q3 and L12 leads to

0 | 4.0. |

q q

1. With rightwards positive, the net force on q3 is

12. As a result of the first action, both sphere W and sphere A possess charge 1

2qA, where qAis the initial charge of sphere A. As a result of the second action, sphere W has charge

As a result of the final action, sphere W now has charge equal to

Aq e e

Setting this final expression equal to +18e as required by the problem leads (after a couple of algebra steps) to the answer: qA=+16e.

q q F k

(b) On the right, a force diagram is shown as well as our choice of y axis (the dashed line).

They axis is meant to bisect the line between q2 and q3 in order to make use of the symmetry in the problem (equilateral triangle of side length d, equal-magnitude charges q1 = q2 = q3 = q). We see that the resultant force is along this symmetry axis, and we obtain

y q F k

14. (a) According to the graph, when q3 is very close to q1 (at which point we can consider the force exerted by particle 1 on 3 to dominate) there is a (large) force in the positive x direction. This is a repulsive force, then, so we conclude q1 has the same sign asq3. Thus, q3 is a positive-valued charge.

(b) Since the graph crosses zero and particle 3 is between the others, q1 must have the same sign as q2, which means it is also positive-valued. We note that it crosses zero at r

= 0.020 m (which is a distance d = 0.060 m from q2). Using Coulomb’s law at that point, we have q q q q d q q q q

15. (a) There is no equilibrium position for q3between the two fixed charges, because it is being pulled by one and pushed by the other (since q1 and q2 have different signs); in this region this means the two force arrows on q3 are in the same direction and cannot cancel. It should also be clear that off-axis (with the axis defined as that which passes through the two fixed charges) there are no equilibrium positions. On the semi-infinite region of the axis which is nearest q2 and furthest from q1 an equilibrium position for q3 cannot be found because |q1| < |q2| and the magnitude of force exerted by q2 is everywhere (in that region) stronger than that exerted by q1 on q3. Thus, we must look in the semi-infinite region of the axis which is nearest q1 and furthest from q2, where the net force on q3 has magnitude q q q q k k

withL = 10 cm and 0L isassumed to be positive. We set this equal to zero, as required by the problem, and cancel k and q3. Thus, we obtain q q L L q

which yields (after taking the square root)

for the distance between q3 and q1.That is, 3q should be placed at 14 cmx along the x-axis.

(b) As stated above, y = 0.

16. Since the forces involved are proportional to q, we see that the essential difference between the two situations is FavqB + qC (when those two charges are on the same side) versusFbv qB + qC (when they are on opposite sides). Setting up ratios, we have

= qB + qC

q q

After noting that the ratio on the left hand side is very close to – 7, then, after a couple of algebra steps, we are led to

17. (a) The distance between q1 and q2 is

The magnitude of the force exerted by q1 on q2 is q q F k

(b) The vector G F21 is directed towards q1 and makes an angle T with the +x axis, where y y

(c) Let the third charge be located at (x3,y3), a distance r from q2. We note that q1,q2 and q3 must be collinear; otherwise, an equilibrium position for any one of them would be impossible to find. Furthermore, we cannot place q3 on the same side of q2 where we also findq1, since in that region both forces (exerted on q2 by q3 and q1) would be in the same direction (since q2 is attracted to both of them). Thus, in terms of the angle found in part

(a), we have x3 = x2 – r cosT and y3 = y2 – r sinT (which means y3 > y2 since T is negative).

The magnitude of force exerted on q2 by q3 is 2 2323||Fkqqr , which must equal that of the force exerted on it by q1 (found in part (a)). Therefore,

18. (a) For the net force to be in the +x direction, the y components of the individual forces must cancel. The angle of the force exerted by the q1 = 40 PC charge on 320qCP is 45°, and the angle of force exerted on q3 by Q is at –T where

Therefore, cancellation of y components requires q q Q q kkTq

from which we obtain |Q| = 83 PC. Charge Q is “pulling” on q3, so (since q3 > 0) we concludeQ = –83 PC.

(b) Now, we require that the x components cancel, and we note that in this case, the angle of force on q3 exerted by Q is +T (it is repulsive, and Q is positive-valued). Therefore, cos45 cos q q Qq kkTq

from which we obtain Q = 5.2 PC 5 CP|.

q q F

The signs are chosen so that a negative force value would cause q to move leftward. We requireFq = 0 and solve for q3:

qqx q q

wherex = L/3 is used. Note that we may easily verify that the force on 4.00q also vanishes:

q q q q F

19. (a) If the system of three charges is to be in equilibrium, the force on each charge must be zero. The third charge q3 must lie between the other two or else the forces acting on it due to the other charges would be in the same direction and q3 could not be in equilibrium. Suppose q3 is at a distance x from q, and L – x from 4.00q. The force acting on it is then given by q q F where the positive direction is rightward. We require F3 = 0 and solve for x. Canceling common factors yields 1/x2 = 4/(L – x)2 and taking the square root yields 1/x = 2/(L – x).

The solution is x = L/3. With L = 9.0 cm, we have x = 3.0 cm.

(b) Similarly, the y coordinate of q3 is y = 0. (c) The force on q is

2 3, so that the dashed line distance in the figure is

2/3rd | We net force on q1 due to the two charges q3 and q4 (with |q3| = |q4| = 1.60 u |

10 19 C) on the y axis has magnitude q q q q rdSHSHq .

This must be set equal to the magnitude of the force exerted on q1 by q2 = 8.0 u 10

= 5.0 |q3| in order that its net force be zero:

Givend = 2.0 cm, this then leads to D = 1.92 cm.

(b) As the angle decreases, its cosine increases, resulting in a larger contribution from the charges on the yaxis. To offset this, the force exerted by q2 must be made stronger, so that it must be brought closer to q1 (keep in mind that Coulomb’s law is inversely proportional to distance-squared). Thus, D must be decreased.

We note that, due to the symmetry in the problem, there is no y component to the net force on the third particle. Thus, F represents the magnitude of force exerted by q1 or q2

3/2 |

(a) To find where the force is at an extremum, we can set the derivative of this expression equal to zero and solve for x, but it is good in any case to graph the function for a fuller understanding of its behavior – and as a quick way to see whether an extremum point is a maximum or a miminum. In this way, we find that the value coming from the derivative procedure is a maximum (and will be presented in part (b)) and that the minimum is found at the lower limit of the interval. Thus, the net force is found to be zero at x = 0, which is the smallest value of the net force in the interval 5.0 m txt 0.

(b) The maximum is found to be at x = d/2 or roughly 12 cm.

(c) The value of the net force at x = 0 is Fnet= 0.

(d) The value of the net force at x = d/2 is Fnet= 4.9u10

21. If T is the angle between the force and the x-axis, then cosT = x

Of course, this could also be figured as illustrated in part (a), looking at the maximum force ratio by itself and solving, or looking at the minimum force ratio (¾) at T = 180º and solving for [.

2. We note that the problem is examining the force on charge A, so that the respective distances (involved in the Coulomb force expressions) between B and A, and between C andA, do not change as particle B is moved along its circular path. We focus on the endpoints (T = 0º and 180º) of each graph, since they represent cases where the forces (on A) due to B and C are either parallel or antiparallel (yielding maximum or minimum force magnitudes, respectively). We note, too, that since Coulomb’s law is inversely proportional to r² then the (if, say, the charges were all the same) force due to C would be one-fourth as big as that due to B (since C is twice as far away from A). The charges, it

divided by the charge of B), as well as the aforementioned ¼ factor | That is, the force |

turns out, are not the same, so there is also a factor of the charge ratio [ (the charge of C exerted by C is, by Coulomb’s law equal to ±¼[ multiplied by the force exerted by B.

(a) The maximum force is 2F0 and occurs when T = 180º (B is to the left of A, while C is the right of A). We choose the minus sign and write

One way to think of the minus sign choice is cos(180º) = –1. This is certainly consistent with the minimum force ratio (zero) at T = 0º since that would also imply

(b) The ratio of maximum to minimum forces is 1.25/0.75 = 5/3 in this case, which implies

23. The charge dq within a thin shell of thickness dr is dqdVAdrUU where A = 4Sr2 .

Thus, with U = b/r, we have

q dq b rdr b r r

24. The magnitude of the force is

F k

N mC C m c h c h

25. (a) The magnitude of the force between the (positive) ions is given by

F q q r k qr b gb g

whereq is the charge on either of them and r is the distance between them. We solve for the charge:

q r

N m C

(b) Let nbe the number of electrons missing from each ion. Then, ne = q, or q n

26. Keeping in mind that an Ampere is a Coulomb per second (1 A = 1 C/s), and that a minute is 60 seconds, the charge (in absolute value) that passes through the chest is

This charge consists of n electrons (each of which has an absolute value of charge equal toe). Thus,

27. Eq. 21-1 (in absolute value) gives

. N m C C

c h

(b) If n is the number of excess electrons (of charge –e each) on each drop then

29. The unit Ampere is discussed in §21-4. The proton flux is given as 1500 protons per square meter per second, where each proton provides a charge of q = +e. The current m2 would be

19 | m |

s m C proton Ac h c h

30. Since the graph crosses zero, q1 must be positive-valued: q1=+8.00e. We note that it crosses zero at r = 0.40 m. Now the asymptotic value of the force yields the magnitude and sign of q2:

31. The volume of 250 cm3 corresponds to a mass of 250 g since the density of water is

1.0 g/cm3 . This mass corresponds to 250/18 = 14 moles since the molar mass of water is

18. There are ten protons (each with charge q = +e) in each molecule of H2O, so

with the values of the charges (stated in the problem) plugged in. Finding the value of x which minimizes this expression leads to x= ¼ L. Thus, x = 2.0 cm.

(b) Substituting x= ¼ L back into the expression for the net force magnitude and using the standard value for e leads to Fnet = 9.21u10

(Parte **1** de 2)