This preview shows page 1. Sign up to view the full content.
Unformatted text preview: the
distance from the common center of the shells and will be the distance at which we are to
evaluate the electric field.
Because the electric field has spherical symmetry, the magnitude E of the electric field is
constant at all points on any such spherical Gaussian surface. Furthermore, the electric field
is directed either radially outward (if the net charge within the Gaussian surface is positive)
or radially inward (if the net charge within the Gaussian surface is negative). This means
that the angle φ between the electric field and the normal to any such spherical Gaussian
surface is either 0.0° or 180°. The quantity E cos φ , therefore, is constant, and may be
factored out of the summation in Equation 18.6:
Φ E = Σ ( E cos φ ) ∆A = ( E cos φ ) Σ∆A (1) The sum Σ∆A of all the tiny sections of area ∆A that compose a spherical Gaussian surface
is the total surface area Σ∆A = A = 4π r 2 of a sphere of radius r. Thus, Equation (1) becomes
Φ E = ( E cos φ ) Σ∆A = 4π r 2 E cos φ (2) Chapter 18 Problems 993 SOLUTION
a. The outer shell has a radius r2 = 0.15 m, and we are to determine the electric field at a
distance r = 0.20 m from the common center of the shells. Therefore, we will choose a
spherical Gaussian surface (radius r = 0.20 m) that encloses both shells and shares their
common center. According to Gauss’ law and Equation (2), we have that the net electric
flux ΦE through this sphere is
Σ ( E cos φ ) ∆A = 4π r 2 E cos φ = Q (3) ε0 Solving Equation (3) for E yields
E= Q (4) 4πε 0 r 2 cos φ Because the chosen Gaussian surface encloses both shells, the net charge Q enclosed by the
surface is Q = q1 + q2. The positive charge q2 on the outer shell has a larger magnitude than
the negative charge q1 on the inner shell, so that Q is a positive net charge. Therefore, the
electric field is directed radially outward , and the angle between the electric field and the
normal to the surface of the spherical Gaussian surface is φ = 0.0°. Therefore, Equation (4)
gives the electric field magnitude as
E= q1 + q2
Q
−1.6 × 10 −6 C + 5.1× 10 −6 C
=
=
4πε 0 r 2 cos φ 4πε 0 r 2 cos φ 4π 8.85 × 10 −12 C 2 / N ⋅ m 2 ( 0.20 m )2 cos 0.0o ( ) = 7.9 × 105 N/C b. We again choose a spherical Gaussian surface concentric with the shells, this time of
radius r = 0.10 m. The radius of this sphere is greater than the radius (r1 = 0.050 m) of the
inner shell but less than the radius (r2 = 0.15 m) of the outer shell. Therefore, this Gaussian
surface is located between the two shells and encloses only the charge on the inner shell:
Q = q1. This is a negative charge, so that the electric field is directed radially inward , and
the angle between the electric field and the normal to the surface of the Gaussian sphere is
φ = 180°. From Equation (4), then, we have that
E= q1
Q
−1.6 × 10 −6 C
=
=
4πε 0 r 2 cos φ 4πε 0 r 2 cos φ 4π 8.85 × 10 −12 C 2 / N ⋅ m 2 ( 0.10 m )2 cos180o = 1.4 × 106 N/C ( ) 994 ELECTRIC FORCES AND ELECTRIC FIELDS c. Choosing a spherical Gaussian surface with a radius of r = 0.025 m, we see that it is
entirely inside the inner shell (r1 = 0.050 m). Therefore, the enclosed charge is zero:
Q = 0 C. Equation (4) shows that the electric field at this distance from the common center
is zero:
E= Q
0C
=
= 0 N/C
2
4πε 0 r cos φ 4πε 0 r 2 cos φ 61. REASONING We use a Gaussian surface that is a sphere centered within the solid sphere.
The radius r of this surface is smaller than the radius R of the solid sphere. Equation 18.7
gives Gauss’ law as follows:
Q
Σ ( E cos φ ) ∆ A =
(18.7)
14 244 ε 0
4
3
Electric flux, Φ E We will deal first with the left side of this equation and evaluate the electric flux ФE. Then
we will evaluate the net charge Q within the Gaussian surface.
SOLUTION The positive charge is
Normal
spread uniformly throughout the solid
sphere and, therefore, is spherically Angle φ between E
E
symmetric. Consequently, the electric and the normal is 0º
field is directed radially outward, and for
each element of area ∆ A is perpendicular
to the surface. This means that the
angle φ between the normal to the surface and the field is 0º, as the drawing shows.
Furthermore, the electric field has the same magnitude everywhere on the Gaussian surface.
Because of these considerations, we can write the electric flux as follows: Σ ( E cos φ ) ∆ A = Σ ( E cos 0° ) ∆ A = E ( Σ∆ A )
The term Σ∆ A is the entire area of the spherical Gaussian surface or 4πr2. With this
substitution, the electric flux becomes ( Σ ( E cos φ ) ∆ A = E ( Σ∆ A) = E 4π r 2 ) (1) Now consider the net charge Q within the Gaussian surface. This charge is the charge
density times the volume 4 π r 3 encompassed by that surface:
3
Q= ( ) q
qr 3
× 4 π r3 = 3
3
4 π R3
R
32
44
1 3 123
Charge
density Volume of
Gaussian
surface (2) Chapter 18 Problems 995 Substituting Equations (1) and (2) into Equation 18.7 gives ( ) E 4π r 2 = qr 3 / R3...
View
Full
Document
This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

Click to edit the document details