GENERAL PHYSICS 2
Science, Technology, Engineering and Mathematics
GP2_Q3-W4
GAUSS’S LAW
Time
Allotment
1 hour
1 hour
Learning Area
Most Essential
Learning
Competencies
General Physics 1. Calculate
2
electric flux
1. Use Gauss’s law
to infer electric field
due to uniformly
distributed charges
on long wires,
spheres, and large
plates
Learning Tasks
Read Lesson 1 –
Gauss’s Law and
Electric Flux
Read Lesson 2 –
Gauss’s Law and
Electric Fields
Mode of Delivery
Modular Distance
Learning
Personal
submission of the
module by
parent/guardian/h
ousemate to the
class adviser
every Friday
Answer Assignment # 4
Directions: Write your
answer in a yellow
paper. Please provide
detailed explanations in
a conceptual question
and pertinent solutions
in a mathematical
problem. Each number
is worth 20 points.
LESSON 1 – GAUSS’S LAW AND ELECTRIC FLUX
•
•
Electric flux is property of an electric field that may be thought of as the number of electric
field lines of force (or electric field lines) that intersect a given area.
Gauss’s Law states that the electric flux ΦE through a surface is related to the magnitude E
of the electric field, the area A of the surface, and the angle θ that specifies the direction of
the field relative to the normal to the surface.
In presenting Gauss’ law, it will be necessary to introduce a new idea called electric flux. The idea
of flux involves both the electric field and the surface through which it passes. By bringing together
the electric field and the surface through which it passes, we will be able to define electric flux and
then present Gauss’ law.
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We begin by developing a version of Gauss’ law that applies only to a point charge, which we
assume to be positive. The electric field lines for a positive point charge radiate outward in all
directions from the charge. The magnitude E of the electric field at a distance r from the charge is E
= kq/r2 , in which we have replaced the symbol |q| with the symbol q since we are assuming that the
charge is positive. The constant k can be expressed as 𝑘 = 1/(4𝜋𝜖0 ), where 𝜖0 is the permittivity of
free space. With this substitution, the magnitude of the electric
field becomes 𝐸 = 𝑞/(4𝜋𝜖0 𝑟 2 ). We now place this point charge
at the center of an imaginary spherical surface of radius r, as
Figure 1 shows. Such a hypothetical closed surface is called a
Gaussian surface, although in general it need not be spherical.
The surface area A of a sphere is 𝐴 = 4𝜋𝑟 2, and the magnitude
of the electric field can be written in terms of this area as 𝐸 =
𝑞 /(𝐴 𝜖0 ) or
𝑞
𝐸𝐴 =
𝜖0
Where EA = ΦE = electric flux
Figure 1 A positive point charge is located at the center of an
imaginary spherical surface of radius r. Such a surface is one
example of a Gaussian surface. Here the electric field is
perpendicular to the surface and has the same magnitude everywhere on it.
SAMPLE PROBLEM # 1 Electric Flux Through Sphere
What is the electric flux through a sphere that has a radius of 1.00 m and carries a charge of +1.00
µC at its center?
Solution The magnitude of the electric field 1.00 m from this charge is
The field points radially outward and is therefore everywhere perpendicular to the surface of the
sphere. The flux through the sphere (whose surface area A=4πr 2 = 12.6 m2is thus
LESSON 2 – GAUSS’S LAW AND ELECTRIC FIELDS
The Electric Field of a Charged Thin Spherical Shell
Figures 2a and b show a thin spherical shell of radius R (for clarity, only half of the shell is shown).
A positive charge q is spread uniformly over the shell. Find the magnitude of the electric field at any
point (a) outside the shell and (b) inside the shell.
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Reasoning Because the charge is distributed uniformly over the spherical shell, the electric field is
symmetrical. This means that the electric field is directed radially outward in all directions, and its
magnitude is the same at all points that are equidistant from the shell. All such points lie on a sphere,
so the symmetry is called spherical symmetry. With this symmetry in mind, we will use a spherical
Gaussian surface to evaluate the electric flux ΦE. We will then use Gauss’ law to determine the
magnitude of the electric field.
Figure
2
A
uniform
distribution
of
positive
charge resides on a thin
spherical shell of radius R.
The spherical Gaussian
surfaces S and S1 are used
to evaluate the electric flux
(a) outside and (b) inside the
shell,
respectively.
For
clarity, only half the shell and
the Gaussian surfaces are
shown.
Solution (a) To find the magnitude of the electric field outside the charged shell, we evaluate the
electric flux Φ𝐸 = Σ(𝐸 cos 𝜙)Δ𝐴 by using a spherical Gaussian surface of radius r (r > R) that is
concentric with the shell. See the blue surface labeled S in Figure 2a. Since the electric field is
everywhere perpendicular to the Gaussian surface, 𝜙 = 0° 𝑎𝑛𝑑 cos 𝜙 = 1. In addition, E has the
same value at all points on the surface, since they are equidistant from the charged shell. Being
constant over the surface, E can be factored outside the summation, with the result that
The term ΣΔ𝐴 is just the sum of the tiny areas that make up the Gaussian surface. Since the area
of a spherical surface is 4𝜋𝑟 2 we have ΣΔ𝐴 = 4𝜋𝑟 2. Setting the electric flux equal to 𝑄/𝜖0 , as
specified by Gauss’ law, yields 𝐸 (4𝜋𝑟 2 ) = 𝑄/𝜖0. Since the only charge within the Gaussian surface
is the charge q on the shell, it follows that the net charge within the Gaussian surface is 𝑄 = 𝑞.
Thus, we can solve for 𝐸 and find that
This is a surprising result, for it is the same as that for a point charge. Thus, the electric field outside
a uniformly charged spherical shell is the same as if all the charge q were concentrated as a point
charge at the center of the shell.
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(b) To find the magnitude of the electric field inside the charged shell, we select a spherical Gaussian
surface that lies inside the shell and is concentric with it. See the blue surface labeled S 1 in Figure
2b. Inside the charged shell, the electric field (if it exists) must also have spherical symmetry.
Therefore, using reasoning like that in part (a), the electric flux through the Gaussian surface is Φ𝐸 =
Σ(𝐸 cos 𝜙)Δ𝐴 = 𝐸 (4𝜋𝑟1 2 )In accord with Gauss’ law, the electric flux must be equal to 𝑄/𝜖0 , where Q
is the net charge inside the Gaussian surface. But now 𝑄 = 0 𝐶, since all the charge lies on the
𝑄
shell that is outside the surface S1. Consequently, we have, 𝐸 (4𝜋𝑟1 2 ) = 𝜖 = 0, or
0
Gauss’ law allows us to deduce that there is no electric field inside a uniform spherical shell of
charge. An electric field exists only on the outside
The Electric Field Inside a Parallel Plate Capacitor
The electric field inside a parallel plate capacitor, and away
from the edges, is constant and has a magnitude of 𝐸 =
𝜎/𝜖0 , where 𝜎 is the charge density (the charge per unit
area) on a plate. Use Gauss’ law to obtain this result.
Reasoning Figure 3a shows the electric field inside a
parallel plate capacitor. Because the positive and negative
charges are distributed uniformly over the surfaces of the
plates, symmetry requires that the electric field be
perpendicular to the plates. We will take advantage of this
symmetry by choosing our Gaussian surface to be a small
cylinder whose axis is perpendicular to the plates (see part
b of the figure). With this choice, we will be able to evaluate
the electric flux and then, with the aid of Gauss’ law,
determine E.
Figure 3 (a) A side view of a parallel plate capacitor,
showing some of the electric field lines. (b) The Gaussian
surface is a cylinder oriented so its axis is perpendicular to
the positive plate and its left end is inside the plate.
Solution Figure 3b shows that we have placed our Gaussian cylinder so that its left end is inside
the positive metal plate, and the right end is in the space between the plates. To determine the
electric flux through this Gaussian surface, we evaluate the flux through each of the three parts—
labeled 1, 2, and 3 in the drawing—that make up the total surface of the cylinder and then add up
the fluxes.
Surface 1—the flat left end of the cylinder—is embedded inside the positive metal plate. The electric
field is zero everywhere inside a conductor that is in equilibrium under electrostatic conditions. Since
𝐸 = 0 𝑁/𝐶, the electric flux through this surface is also zero:
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Surface 2—the curved wall of the cylinder—is everywhere parallel to the electric field between the
plates, so that cos 𝜙 = cos 90° = 0. Therefore, the electric flux through this surface is also zero:
Surface 3—the flat right end of the cylinder—is perpendicular to the electric field between the plates,
so ccos 𝜙 = cos 0° = 1. The electric field is constant over this surface, so E can be taken outside.
Noting that ΣΔ𝐴 = 𝐴 is the area of surface 3, we find that the electric flux through this surface is
The electric flux through the entire Gaussian cylinder is the sum of the three fluxes determined
above:
According to Gauss’ law, we set the electric flux equal to Q/ 0, where Q is the net charge inside the
Gaussian cylinder: EA Q/ 0. But Q/A is the charge per unit area, , on the plate. Therefore, we arrive
at the value of the electric field inside a parallel plate capacitor: . The distance of the right end of the
Gaussian cylinder from the positive plate does not appear in this result, indicating that the electric
field is the same everywhere between the plate.
ASSIGNMENT # 4
Directions: Write your answer in a yellow paper. Please provide detailed explanations in a
conceptual question and pertinent solutions in a mathematical problem. Each number is worth 20
points.
1. A spherical gaussian surface surrounds a point charge q. Describe what happens to the total
flux through the surface if
(A) the charge is tripled,
(B) the radius of the sphere is doubled,
(C) the surface is changed to a cube, and
(D) the charge is moved to another location inside the surface.
2. A cubical Gaussian surface surrounds two charges, q1 = +6.0 x 10-12 C and q2 = -2.0 x 10-12
C. What is the electric flux passing through the surface?
3. A spherical surface completely surrounds a collection of charges. Find the electric flux
through the surface if the collection consists of (a) a single +3.5 x 10-6 C charge, (b) a single
-2.3 x 10-6 C charge, and (c) both of the charges in (a) and (b).
References:
H. (2022). Conceptual Physics (12th Edition) 12th edition by Hewitt, Paul G. (2014) Hardcover
(12th ed.). Addison-Wesley, 2014.
Johnson, C. (2022). Physics Volume 1, 8th Edition: Department of Physics, PHY 2053 (University
of Central Florida) (8th ed.). Wiley Custom Learning Solutions.
Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers with Modern Physics
(10th ed.). Cengage Learning.
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Submitted by:
Maria Fatima L. Jingco
Subject Teacher
Validated by:
Domingo S. Soriao, Jr.
Master Teacher II
Key to Correction:
Assignment # 3
1.a. answer may vary
2. Φ= 0.452 N.m2/C
3.a. Φ= 0.396 N.m2/C
Ferdinand G. Masangcay
Head Teacher V
b. answer may vary
c. answer may vary
b. Φ= 0.260 N.m2/C
c. Φ= 0.136 N.m2/C
d. answer may vary
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