Resolução Eletromagnetismo, Provas de Eletromagnetismo

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1. Eq. 21-1 gives Coulomb’s Law, F k q^^1 r^2 q 2 , which we solve for the distance:

9 2 2 6 6 | 1 || 2 | 8.99^ 10 N m^ C^ 26.0^10 C^ 47.0^10 C 1.39m. 5.70N

k q q r F

u ˜ u ^ u 

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

m a 2 (^) 2 m a 1 (^) 1 m 2

7 6 3 10 7 0 7 9 0

u u



..  .

kg m s m s

kg.

2 2

c hc h

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

2 1 2 9 2 2 1 1 2 8.99^ 10 N m^ C^ (0.0032 m) 2.

q q q F m a k r

u ˜

Inserting the values for m 1 and a 1 (see part (a)) we obtain | q | = 7.1 u 10 –11^ C.

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

F kq^2 / r^2. 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 3 q /4 in the final situation. The repulsive force between spheres 1 and 2 is finally

2 2 2

q q q F F k k F r r F

c Ÿ

F

q Q q r

4 S H 0 2

b g

where r is the distance between the charges. We want the value of q that maximizes the function f ( q ) = q ( Q – q ). Setting the derivative dF / dq equal to zero leads to Q – 2 q = 0,

or q = Q /2. Thus, q / Q = 0.500.

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

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 q 1 and q 2 be the original charges. We choose the coordinate system so the force on q 2 is positive if it is repelled by q 1. Then, the force on q 2 is

F

q q r

k

q q a (^) r

1 2 2

1 2 S H^2

where r = 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 ( q 1 + q 2 )/2. The force is now one of repulsion and is given by

F

r

k

q q b r

q q q q 

  1 4 0 4

2 2 2

1 2

2 2

1 2 1 2

S H

d id i b g

We solve the two force equations simultaneously for q 1 and q 2. The first gives the product

q q

r F k

a 1 2

2 2 9

u ˜

 u 

m N N m C

2 2 C^2

b g b g

and the second gives the sum

q q r

F

k

b 1 2

 2 2 0 500 0 0360^ 2 00 106

u ˜

. u 

m.

N

8.99 10 N m C

b g 9 2 2 C

where we have taken the positive root (which amounts to assuming q 1 + q 2 t 0). Thus, the product result provides the relation

12 2 2 1

3.00 10 C

q q

 u^ 

which we substitute into the sum result, producing

q q

1

12

1

 3 00^ u^10 2 00 u 106



. C (^).  C. 2

Multiplying by q 1 and rearranging, we obtain a quadratic equation

q 12  c 2 00. u 10 ^6 C h q 1  3 00. u 10 ^12 C^20.

The solutions are

q 1

. u ^ r . u ^  . u  .

C c C h c C^2 h

If the positive sign is used, q 1 = 3.00 u 10 –6^ C, and if the negative sign is used, 6 q 1 (^) 1.00 10 C  u .

(a) Using q 2 = (–3.00 u 10 –12)/ q 1 with q 1 = 3.00 u 10 –6^ C, we get q 2 (^) 1.00 u 10 ^6 C.

(b) If we instead work with the q 1 = –1.00 u 10 –6^ C root, then we find q 2 (^) 3.00 u 10 ^6 C.

Note that since the spheres are identical, the solutions are essentially the same: one sphere

originally had charge –1.00 u 10 –6^ C and the other had charge +3.00 u 10 –6^ C.

What if we had not made the assumption, above, that q 1 + q 2 t 0? If the signs of the charges were reversed (so q 1 + q 2 < 0), then the forces remain the same, so a charge of +1.00 u 10 –6^ C on one sphere and a charge of –3.00 u 10 –6^ C on the other also satisfies the conditions of the problem.

9. The force experienced by q 3 is

3 1 3 2 3 4 (^3 31 32 34 2 ) 0

j (cos45 i sin 45 j) i (^4) ( 2 )

q q q q q q F F F F SH a (^) a a

  (^) ¨   q  q  ¸ © ¹

G G G G

(a) Therefore, the x -component of the resultant force on q 3 is

7 2 3 2 9 2 2 (^3 2 ) 0

| | | | 2 1.0^10 C 1

| | 8.99 10 N m C 2 0.17N. x 4 2 2 (0.050 m) 2 2

q q F q SH a

u^  § · § · ¨ ^ ¸ u^ ˜^ ¨  ¸ © ¹ © ¹

(b) Similarly, the y -component of the net force on q 3 is

7 2 3 2 9 2 2 (^3 2 ) 0

| | | | 2 1.0^10 C 1

| | 8.99 10 N m C 1 0.046N. y 4 2 2 (0.050 m) 2 2

q q F q SH a

u^  § · § · ¨ ^ ^ ¸ u^ ˜^ ¨  ^ ¸  © ¹ © ¹

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

1 2 2 2 0 0

q Q q Q SH  a  a SH a  a

which leads to | q 1 | = 9.0 | q 2 |. Since Q is located between q 1 and q 2 , we conclude q 1 and q 2 are like-sign. Consequently, q 1 / q 2 = 9.0.

(b) Now we have

1 2 2 2 0 0

q Q q Q SH  a  a SH a  a

which yields | q 1 | = 25 | q 2 |. Now, Q is not located between q 1 and q 2 , one of them must push and the other must pull. Thus, they are unlike-sign, so q 1 / q 2 = –25.

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

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

§ q A^  e · ¨ ¸ © ¹

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

ª (^) § q A (^)  e · e º « ¨ ¸ » ¬ ©^ ¹ ¼

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

13. (a) Eq. 21-1 gives

6 2 1 2 9 2 2 (^12 )

20.0 10 C

8.99 10 N m C 1.60N. 1.50m

q q F k d

u^  u ˜

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

The y axis is meant to bisect the line between q 2 and q 3 in order to make use of the symmetry in the problem (equilateral triangle of side length d , equal-magnitude charges q 1 = q 2 = q 3 = q ). We see that the resultant force is along this symmetry axis, and we obtain

2 6 2 9 2 2 2 2

20.0 10 C

2 cos 30 2 8.99 10 N m C cos 30 2.77 N 1.50m

y

q F k d

u^  § · ¨ ¸ q^ u^ ˜^ q © ¹

15. (a) There is no equilibrium position for q 3 between the two fixed charges, because it is being pulled by one and pushed by the other (since q 1 and q 2 have different signs); in this region this means the two force arrows on q 3 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 q 2 and furthest from q 1 an equilibrium position for q 3 cannot be found because | q 1 | < | q 2 | and the magnitude of force exerted by q 2 is everywhere (in that region) stronger than that exerted by q 1 on q 3. Thus, we must look in the semi-infinite region of the axis which is nearest q 1 and furthest from q 2 , where the net force on q 3 has magnitude

1 3 2 3 2 2 (^0 )

q q q q k k L (^) L L

with L = 10 cm and L 0 is assumed to be positive. We set this equal to zero, as required by

the problem, and cancel k and q 3. Thus, we obtain

2 (^1 2 0 ) 2 2 0 0 0 1

3.0 C

1.0 C

q q (^) L L q L (^) L L L q

P P

which yields (after taking the square root)

0 0 0

10 cm 3 14 cm 3 1 3 1

L L L

L

L

for the distance between q 3 and q 1. That is, q 3 should be placed at x 14 cm 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 Fa v qB + qC (when those two charges are on the same side)

versus Fb v  qB + qC (when they are on opposite sides). Setting up ratios, we have

Fa Fb =^

qB + qC

• qB + qC^ Ÿ

23 24

2.014 10 N 1 /

2.877 10 N 1 /

C B C B

q q q q

 

u   u  

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 7 1 8 1.333. 7 1 6

C B

q q

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 q 1 = 40 PC charge on

q 3 (^) 20 P C is 45°, and the angle of force exerted on q 3 by Q is at – T where

tan 1 2.0 cm 33.. 3.0 cm

T ^ §¨^ ·¸^ q © ¹

Therefore, cancellation of y components requires

1 3 3 2 2 2 2

sin 45 sin 0.02 2 m (^) (0.030 m) (0.020 m)

q q Q q k q k T 

from which we obtain | Q | = 83 PC. Charge Q is “pulling” on q 3 , so (since q 3 > 0) we conclude Q = –83 PC.

(b) Now, we require that the x components cancel, and we note that in this case, the angle of force on q 3 exerted by Q is + T (it is repulsive, and Q is positive-valued). Therefore,

1 3 3 2 cos 45^2 22 cos 0.02 2 m (^) (0.030 m) (0.020 m)

q q Qq k q k T 

from which we obtain Q = 55.2 PC | 55 PC.

2 3 2 2 0

q 4

qq q F H x L

 §^ ·

S © ¹

The signs are chosen so that a negative force value would cause q to move leftward. We require Fq = 0 and solve for q 3 :

2 3 (^3 )

qx q q q L q

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

2 2 2 2 2 0 (^4 2 22 2 2 ) 0 0 0

q 4 4 4 9 4

q qq q q q q F H L (^) L x H L L H L L

S © ¹ S S

19. (a) If the system of three charges is to be in equilibrium, the force on each charge must be zero. The third charge q 3 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 q 3 could not be in equilibrium. Suppose q 3 is at a distance x from q , and L – x from 4.00 q. The force acting on it is then given by

3 3 3 2 2 0

qq qq F SH x (^) L x

where the positive direction is rightward. We require F 3 = 0 and solve for x. Canceling common factors yields 1/ x^2 = 4/( L – x )^2 and taking the square root yields 1/ x = 2/( L – x ). The solution is x = L /3. With L = 9.00 cm, we have x = 3.00 cm.

(b) Similarly, the y coordinate of q 3 is y = 0.

(c) The force on q is