**Solução Halliday Volume 3 - Eletromagnetismo - ch32**

Resolução do Halliday - Volume 3 - Eletromagnetismo

(Parte **1** de 2)

1. We note that the symbol q2 is used in the problem statement to mean the absolute value of the negative charge which resides on the larger shell. The following sketch is for

12qq=.

The following two sketches are for the cases q1 > q2 (left figure) and q1 < q2 (right figure).

2. (a) We note that the electric field points leftward at both points. UsingGG FqE=0, and orienting our x axis rightward (so i points right in the figure), we find

C iNich which means the magnitude of the force on the proton is 6.4 × 10–18 N and its direction

(b) As the discussion in §22-2 makes clear, the field strength is proportional to the “crowdedness” of the field lines. It is seen that the lines are twice as crowded at A than at

B, so we conclude that EA = 2EB. Thus, EB = 20 N/C.

3. The following diagram is an edge view of the disk and shows the field lines above it. Near the disk, the lines are perpendicular to the surface and since the disk is uniformly charged, the lines are uniformly distributed over the surface. Far away from the disk, the lines are like those of a single point charge (the charge on the disk). Extended back to the disk (along the dotted lines of the diagram) they intersect at the center of the disk.

If the disk is positively charged, the lines are directed outward from the disk. If the disk is negatively charged, they are directed inward toward the disk. A similar set of lines is associated with the region below the disk.

4. We find the charge magnitude |q| from E = |q|/4πε0r2 :

5. Since the magnitude of the electric field produced by a point charge q is given by

20||/4Eqrπε=, where r is the distance from the charge to the point where the field has magnitude E, the magnitude of the charge is

6. With x1 = 6.0 cm and x2 = 21.0 cm, the point midway between the two charges is located at x = 13.5 cm. The values of the charge are q1 = –q2 = – 2.0 × 10–7 C, and the magnitudes and directions of the individual fields are given by:

q E

q E xπε πε

Thus, the net electric field is

7. Since the charge is uniformly distributed throughout a sphere, the electric field at the surface is exactly the same as it would be if the charge were all at the center. That is, the magnitude of the field is

E q

where q is the magnitude of the total charge and R is the sphere radius. (a) The magnitude of the total charge is Ze, so

E Ze

.. Nm C C

NC c hb gc h

c h

(b) The field is normal to the surface and since the charge is positive, it points outward from the surface.

8. (a) The individual magnitudes G

E1 and G E2 are figured from Eq. 2-3, where the absolute value signs for q2 are unnecessary since this charge is positive. Whether we add the magnitudes or subtract them depends on if G E1 is in the same, or opposite, direction as

E2. At points left of q1 (on the –x axis) the fields point in opposite directions, but there is no possibility of cancellation (zero net field) since G

E1 is everywhere bigger than G E2 in this region. In the region between the charges (0 < x < L) both fields point leftward and there is no possibility of cancellation. At points to the right of q2 (where x > L), G E1 points leftward and G E2 points rightward so the net field in this range is

Although |q1| > q2 there is the possibility of G Enet=0 since these points are closer to q2 than to q1. Thus, we look for the zero net field point in the x > L region:

|||| |

q E

which leads to

xL=≈− .

(b) A sketch of the field lines is shown in the figure below:

9. At points between the charges, the individual electric fields are in the same direction and do not cancel. Since charge q2= − 4.0 q1 located at x2 = 70 cm has a greater magnitude than q1 = 2.1 ×10 −8 C located at x1 = 20 cm, a point of zero field must be closer to q1 than to q2. It must be to the left of q1.

Let x be the coordinate of P, the point where the field vanishes. Then, the total electric field at P is given by q E

x xπε

If the field is to vanish, then q q x x

x qx x

Choosing –2.0 for consistency, the value of x is found to be x = −30 cm.

10. We place the origin of our coordinate system at point P and orient our y axis in the direction of the q4 = –12q charge (passing through the q3 = +3q charge). The x axis is perpendicular to the y axis, and thus passes through the identical q1 = q2 = +5q charges.

EEE123 and ||G E4 are figured from Eq. 2-3, where the absolute value signs for q1, q2, and q3 are unnecessary since those charges are positive (assuming q > 0). We note that the contribution from q1 cancels that of q2 (that is,

EE12=), and the net field (if there is any) should be along the y axis, with magnitude equal to

qd qd qd q d net j=− which is seen to be zero. A rough sketch of the field lines is shown below:

1. The x component of the electric field at the center of the square is given by q q E

q q q

Similarly, the y component of the electric field is q q E q q q

12. By symmetry we see the contributions from the two charges q1 = q2 = +e cancel each other, and we simply use Eq. 2-3 to compute magnitude of the field due to q3 = +2e.

(a) The magnitude of the net electric field is e E

ra aπε πε πε −−

(b) This field points at 45.0°, counterclockwise from the x axis.

13. (a) The vertical components of the individual fields (due to the two charges) cancel, by symmetry. Using d = 3.0 m, the horizontal components (both pointing to the –x direction) add to give a magnitude of

Ex, net = 2 q d

−10 N/C .

(b) The net electric field points in the –x direction, or 180° counterclockwise from the +x axis.

(caused by q1 and q2) must point in opposite directions for x > 0. Given their locations in the figure, we conclude they are therefore oppositely charged. Further, since the net field points more strongly leftward for the small positive x (where it is very close to q2) then we conclude that q2 is the negative-valued charge. Thus, q1 is a positive-valued charge. We write each charge as a multiple of some positive number ξ (not determined at this point). Since the problem states the absolute value of their ratio, and we have already inferred their signs, we have q1 = 4 ξ and q2 = −ξ. Using Eq. 2-3 for the individual fields, we find

Enet = E1 + E2 =

4πεo (L + x) 2 –

4πεo x 2

leads to L = 20 cm |

for points along the positive x axis. Setting Enet = 0 at x = 20 cm (see graph) immediately

is maximum), we obtain (after some simplification) that location: |

(a) If we differentiate Enet with respect to x and set equal to zero (in order to find where it

We note that the result for part (a) does not depend on the particular value of ξ.

(b) Now we are asked to set ξ = 3e, where e = 1.60 ×10 −19

C, and evaluate Enet at the value of x (converted to meters) found in part (a). The result is 2.2 × 10 −8 N/C .

14. For it to be possible for the net field to vanish at some x > 0, the two individual fields

6 | m NC |

The directions are indicated in standard format below. We use the magnitude-angle notation (convenient if one is using a vector-capable calculator in polar mode) and write

(starting with the proton on the left and moving around clockwise) the contributions to G Enet as follows:

(a) The result above shows that the magnitude of the net electric field is

(b) Similarly, the direction of G Enet is –76.4° from the x axis.

15. The field of each charge has magnitude

16. The net field components along the x and y axes are

Enet x = q1 q2 cos θ

2 , | Enet y = – |

4πεoR q2 sin θ

2 |

4πεoR

The magnitude is the square root of the sum of the components-squared. Setting the magnitude equal to E = 2.0 × 105 N/C, squaring and simplifying, we obtain

4 |

16π2εo2 R

With R = 0.50 m, q1 = 2.0 × 10 − 6

C and q2 = 6.0 × 10

− 6 C, we can solve this expression for cos θ and then take the inverse cosine to find the angle. There are two answers.

(a) The positive value of angle is θ = 67.8°. (b) The positive value of angle is θ = − 67.8°.

17. The magnitude of the dipole moment is given by p = qd, where q is the positive charge in the dipole and d is the separation of the charges. For the dipole described in the problem,

−−−16010430106881019928 | C m Cmchch. |

18. According to the problem statement, Eact is Eq. 2-5 (with z = 5d)

9801 πεo d 2 and Eapprox is q d

2 |

250πεo d

The ratio is therefore

Eapprox

Eact = 0.9801 ≈ 0.98.

(a) The magnitude of the net electric field at point P is qd qd E

dr dr drθ πε πε qd E rπε ≈G

(b) From the figure, it is clear that the net electric field at point P points in the − j direction, or −90° from the +x axis.

19. Consider the figure below.

20. Referring to Eq. 2-6, we use the binomial expansion (see Appendix E) but keeping higher order terms than are shown in Eq. 2-7:

E = q

z 3 + …

= |

q d

2πεo z 3 + q d 3

4πεo z 5 + … magnitude p = qd. The moments point in opposite directions and produce fields in opposite directions at points on the quadrupole axis. Consider the point P on the axis, a distance z to the right of the quadrupole center and take a rightward pointing field to be positive. Then, the field produced by the right dipole of the pair is qd/2πε0(z – d/2)3 and

qd z dzz dz qd z

0 4π πε ε .

Let Q = 2qd 2 . Then,

21. Think of the quadrupole as composed of two dipoles, each with dipole moment of

2. We use Eq. 2-3, assuming both charges are positive. At P, we have qR q R E

23. (a) We use the usual notation for the linear charge density: λ = q/L. The arc length is L = rθ if θ is expressed in radians. Thus,

L = (0.0400 m)(0.698 rad) = 0.0279 m.

−15 C/m.

(b) We consider the same charge distributed over an area A = πr2 = π(0.0200 m)2 and

(c) Now the area is four times larger than in the previous part (Asphere = 4πr2 ) and thus obtain an answer that is one-fourth as big:

−15 C/m².

(d) Finally, we consider that same charge spread throughout a volume of 4π r3 /3 and semicircular charge +=⋅qRλπ(and that it points downward). Adapting the steps leading to Eq. 2-21, we find q E

(b) The net electric field netEG points in the j−direction, or 90−°counterclockwise from the +x axis.

24. From symmetry, we see that the net field at P is twice the field caused by the upper

25. Studying Sample Problem 2-4, we see that the field evaluated at the center of curvature due to a charged distribution on a circular arc is given by

/ along the symmetry axis

where λ = q/rθ with θ in radians. In this problem, each charged quarter-circle produces a field of magnitude q E

r rθ επ ε π−

That produced by the positive quarter-circle points at – 45°, and that of the negative quarter-circle points at +45°.

(a) The magnitude of the net field is q E rπε π πε π π −−

(b) By symmetry, the net field points vertically downward in the j−direction, or 90−° counterclockwise from the +x axis.

26. We find the maximum by differentiating Eq. 2-16 and setting the result equal to zero.

ddz qzzR qR z zR4 4 which leads tozR=/2. With R = 2.40 cm, we have z = 1.70 cm.

27. (a) The linear charge density is the charge per unit length of rod. Since the charge is uniformly distributed on the rod,

0.0815 mqL

(b) We position the x axis along the rod with the origin at the left end of the rod, as shown in the diagram.

Let dx be an infinitesimal length of rod at x. The charge in this segment isdqdx=λ. The charge dq may be considered to be a point charge. The electric field it produces at point P has only an x component and this component is given by dE dx

La x

The total electric field produced at P by the whole rod is the integral dx E

La x a LaLa x

Lq aL a a L a upon substituting qLλ−=. With q = 4.23 × 10 −15 C, L =0.0815 m and a = 0.120 m, we

(c) The negative sign indicates that the field points in the –x direction, or −180° counterclockwise form the +x axis.

(d) If a is much larger than L, the quantity L + a in the denominator can be approximated by a and the expression for the electric field becomes

E q

Since 50 m 0.0815 m,aL== the above approximation applies and we have

(e) For a particle of charge 154.2310 C,q− −=−×the electric field at a distance a = 50 m

28. First, we need a formula for the field due to the arc. We use the notation λ for the charge density, λ = Q/L. Sample Problem 2-4 illustrates the simplest approach to circular arc field problems. Following the steps leading to Eq. 2-21, we see that the general result (for arcs that subtend angle θ) is

Earc = λ

2πεo r

Now, the arc length is L = rθ if θ is expressed in radians. Thus, using R instead of r, we obtain

Earc = Q/L sin(θ/2)

2πεo R = Q sin(θ/2)

2 |

2πεo θ R

Thus, with θ = π, the problem asks for the ratio Eparticle / Earc where Eparticle is given by Eq. 2-3. We obtain

on an element dx of the rod contains charge dq = λ dx. By symmetry, we conclude that all horizontal field components (due to the dq’s) cancel and we need only “sum” (integrate) the vertical components. Symmetry also allows us to integrate these contributions over only half the rod (0 ≤ x ≤ L/2) and then simply double the result. In that regard we note that sin θ = R/r where 2 rxR=+.

(a) Using Eq. 2-3 (with the 2 and sin θ factors just discussed) the magnitude is

2 2s in dq dx y E rx R xR

qL RRd x x Rx RxR qL q where the integral may be evaluated by elementary means or looked up in Appendix E (item #19 in the list of integrals). With 127.8110 Cq−=×, 0.145 mL=and R = 0.0600 m,

(b) As noted above, the electric fieldEG points in the +y direction, or

29. We assume q > 0. Using the notation λ = q/L we note that the (infinitesimal) charge

30. From Eq. 2-26 z E

31. At a point on the axis of a uniformly charged disk a distance z above the center of the disk, the magnitude of the electric field is

E z

where R is the radius of the disk and σ is the surface charge density on the disk. See Eq.

2-26. The magnitude of the field at the center of the disk (z = 0) is Ec = σ/2ε0. We want to solve for the value of z such that E/Ec = 1/2. This means zzR z zR

Squaring both sides, then multiplying them by z2 + R2 , we obtain z2 = (z2 /4) + (R2

/4).

Thus, z2 = R2 /3, or zR=3. With R = 0.600 m, we have z = 0.346 m.

32. We write Eq. 2-26 as

Emax = 1 – and note that this ratio is 1

2 (according to the graph shown in the figure) when z = 4.0 cm.

Solving this for R we obtain R = z 3 = 6.9 cm.

disk in figure (a) plus a concentric smaller disk (of radius R/2) with the opposite value of σ. That is,

where

2 |

We find the relative difference and simplify:

4 + ¼

= 0.283 or approximately 28%.

3. We use Eq. 2-26, noting that the disk in figure (b) is effectively equivalent to the qE mg E mge

(b) Since the force of gravity is downward, then qEG must point upward. Since q > 0 in this situation, this implies G E must itself point upward.

34. (a) Vertical equilibrium of forces leads to the equality

35. The magnitude of the force acting on the electron is F = eE, where E is the magnitude of the electric field at its location. The acceleration of the electron is given by Newton’s second law:

Fm eE m ms 4 2c hc h

36. Eq. 2-28 gives

Fq mae m e a== −

=−FHGIKJbg using Newton’s second law.

(a) With east being the i direction, we have

C msiNCi2ej which means the field has a magnitude of 0.0102 N/C

(b) The result shows that the fieldEG is directed in the –x direction, or westward.

37. We combine Eq. 2-9 and Eq. 2-28 (in absolute values).

Fq Eq p z kep z where we have used Eq. 21-5 for the constant k in the last step. Thus, we obtain c hc hc h

c h which yields a force of magnitude 6.6 × 10–15 N. If the dipole is oriented such that G p is in the +z direction, then G F points in the –z direction.

(Parte **1** de 2)