This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ε0 Rearranging this result shows that E= qr 3 / R3 ( 4π r ) ε 0
2 = qr
4πε 0 R3 62. REASONING Because the charge is distributed uniformly along the straight wire, the
electric field is directed radially outward, as the following end view of the wire illustrates.
Long straight wire Gaussian cylinder E Gausssian
cylinder r +++ + + +++
2 1 3 L
End view And because of symmetry, the magnitude of the electric field is the same at all points
equidistant from the wire. In this situation we will use a Gaussian surface that is a cylinder
concentric with the wire. The drawing shows that this cylinder is composed of three parts,
the two flat ends (1 and 3) and the curved wall (2). We will evaluate the electric flux for
this threepart surface and then set it equal to Q/ε0 (Gauss’ law) to find the magnitude of the
electric field.
SOLUTION Surfaces 1 and 3 – the flat ends of the cylinder – are parallel to the electric
field, so cos φ = cos 90° = 0. Thus, there is no flux through these two surfaces:
Φ1 = Φ3 = 0 N ⋅ m 2 /C .
Surface 2 – the curved wall – is everywhere perpendicular to the electric field E, so
cos φ = cos 0° = 1. Furthermore, the magnitude E of the electric field is the same for all
points on this surface, so it can be factored outside the summation in Equation 18.6: Φ 2 = Σ (E cos 0° )∆ A = E Σ A
The area ΣA of this surface is just the circumference 2π r of the cylinder times its length L:
ΣA = (2π r)L. The electric flux through the entire cylinder is, then, 996 ELECTRIC FORCES AND ELECTRIC FIELDS Φ E = Φ 1 + Φ 2 + Φ 3 = 0 + E (2 π rL ) + 0 = E (2 π rL )
Following Gauss’ law, we set ΦE equal to Q/ε0, where Q is the net charge inside the Gaussian cylinder: E(2π rL) = Q/ε0. The ratio Q/L is the charge per unit length of the wire
and is known as the linear charge density λ: λ = Q/L. Solving for E, we find that
E= Q/L
λ
=
2π ε 0 r 2π ε 0 r ______________________________________________________________________________
63. REASONING AND SOLUTION The electric
field lines must originate on the positive charges
and terminate on the negative charge. They
3q
cannot cross one another. Furthermore, the
number of field lines beginning or terminating on
any charge must be proportional to the magnitude
of the charge. Thus, for every field line that
leaves the charge +q, two field lines must leave
the charge +2q. These three lines must terminate
+q
+2q
on the −3q charge. If the sketch is to have six
field lines, two of them must originate on +q, and
four of them must originate on the charge +2q.
______________________________________________________________________________
64. REASONING
a. The magnitude of the electrostatic force that acts on each sphere is given by Coulomb’s
law as F = k q1 q2 / r 2 , where q1 and q2 are the magnitudes of the charges, and r is the
distance between the centers of the spheres.
b. When the spheres are brought into contact, the net charge after contact and separation
must be equal to the net charge before contact. Since the spheres are identical, the charge
on each after being separated is onehalf the net charge. Coulomb’s law can be applied
again to determine the magnitude of the electrostatic force that each sphere experiences. SOLUTION
a. The magnitude of the force that each sphere experiences is given by Coulomb’s law as: F= k q1 q2
r 2 (8.99 × 109 N ⋅ m2 /C2 ) ( 20.0 × 10−6 C) (50.0 × 10−6 C) =
=
2
( 2.50 × 10−2 m ) Because the charges have opposite signs, the force is attractive . 1.44 × 104 N Chapter 18 Problems 997 b. The net charge on the spheres is −20.0 µ C + 50.0 µ C = +30.0 µ C. When the spheres are
brought into contact, the net charge after contact and separation must be equal to the net
charge before contact, or +30.0 µ C. Since the spheres are identical, the charge on each after
being separated is onehalf the net charge, so q1 = q2 = + 15.0 µ C . The electrostatic force
that acts on each sphere is now F= k q1 q2
r 2 (8.99 × 109 N ⋅ m2 /C2 ) (15.0 × 10−6 C ) (15.0 × 10−6 C ) =
=
2
( 2.50 × 10−2 m ) 3.24 × 103 N Since the charges now have the same signs, the force is repulsive .
______________________________________________________________________________
65. REASONING AND SOLUTION The +2q of charge initially on the sphere lies entirely on
the outer surface. When the +q charge is placed inside of the sphere, then a −q charge will
still be induced on the interior of the sphere. An additional +q will appear on the outer
surface, giving a net charge of +3q .
______________________________________________________________________________
66. REASONING Two forces act on the charged ball (charge q); they are the downward force
of gravity mg and the electric force F due to the presence of the charge q in the electric field
E. In order for the ball to float, these two forces must be equal in magnitude and opposite in
direction, so that the net force on the ball is zero (Newton's second law). Therefore, F must
point upward, which we will take as the positive direction. According to Equati...
View Full
Document
 Spring '13
 CHASTAIN
 Physics, The Lottery

Click to edit the document details