Summary 738 Chapter 24 Gauss's Law

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Summary 738 Chapter 24 Gauss's Law
Chapter 24
Gauss’s Law
Chapter 24
Gauss's Law
Electric flux is proportional to the number of electric field lines that penetrate a surface. If the electric field is
uniform and makes an angle u with the normal to a surface of area A, the electric flux through the surface is
In general, the electric flux through a surface is
FE 5 EA cos u
FE ; 3 E ? d A
Concepts and Principles
Gauss’s law says that the net
electric flux FE through any closed
gaussian surface is equal to the net
charge q in inside the surface divided
by P0:
q in
FE 5 C E ? d A 5
Using Gauss’s law, you can calculate
the electric field due to various symmetric charge distributions.
Objective Questions
A conductor in electrostatic equilibrium has the following properties:
1. The electric field is zero everywhere inside the conductor, whether
the conductor is solid or hollow.
2. If the conductor is isolated and carries a charge, the charge
resides on its surface.
3. The electric field at a point just outside a charged conductor is
perpendicular to the surface of the conductor and has a magnitude s/P0, where s is the surface charge density at that point.
4. On an irregularly shaped conductor, the surface charge density is
greatest at locations where the radius of curvature of the surface
is smallest.
1. denotes answer available in Student Solutions Manual/Study Guide
1. A cubical gaussian surface surrounds a long, straight,
charged filament that passes perpendicularly through
two opposite faces. No other charges are nearby.
(i) Over how many of the cube’s faces is the electric
field zero? (a) 0 (b) 2 (c) 4 (d) 6 (ii) Through how many
of the cube’s faces is the electric flux zero? Choose
from the same possibilities as in part (i).
2. A coaxial cable consists of a long, straight filament
surrounded by a long, coaxial, cylindrical conducting
shell. Assume charge Q is on the filament, zero net
charge is on the shell, and the electric field is E1 ^i at
a particular point P midway between the filament and
the inner surface of the shell. Next, you place the cable
into a uniform external field 2E ^i. What is the x component of the electric field at P then? (a) 0 (b) between
0 and E 1 (c) E 1 (d) between 0 and 2E 1 (e) 2E 1
3. In which of the following contexts can Gauss’s law not
be readily applied to find the electric field? (a) near a
long, uniformly charged wire (b) above a large, uniformly charged plane (c) inside a uniformly charged
ball (d) outside a uniformly charged sphere (e) Gauss’s
law can be readily applied to find the electric field in
all these contexts.
4. A particle with charge q is located inside a cubical
gaussian surface. No other charges are nearby. (i) If
the particle is at the center of the cube, what is the
flux through each one of the faces of the cube? (a) 0
(b) q/2P0 (c) q/6P0 (d) q/8P0 (e) depends on the size of
the cube (ii) If the particle can be moved to any point
within the cube, what maximum value can the flux
through one face approach? Choose from the same
possibilities as in part (i).
5. Charges of 3.00 nC, 22.00 nC, 27.00 nC, and 1.00 nC
are contained inside a rectangular box with length
1.00 m, width 2.00 m, and height 2.50 m. Outside the
box are charges of 1.00 nC and 4.00 nC. What is the
electric flux through the surface of the box? (a) 0
(b) 25.64 3 102 N ? m2/C (c) 21.47 3 103 N ? m2/C
(d) 1.47 3 103 N ? m2/C (e) 5.64 3 102 N ? m2/C
6. A large, metallic, spherical shell has no net charge. It
is supported on an insulating stand and has a small
hole at the top. A small tack with charge Q is lowered
on a silk thread through the hole into the interior of
the shell. (i) What is the charge on the inner surface
of the shell, (a) Q (b) Q /2 (c) 0 (d) 2Q /2 or (e) 2Q?
Choose your answers to the following questions from
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