1. Week # 10
Gauss’s Law: Electric charge & electric flux, calculating electric flux,
Gauss’s law, applications of Gauss’s law
Book: “University Physics” by Young & Freedman 13th edition, page: P-725”
Physics-II (Ph-1002), 4 (3, 3)
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2. 22.1 Charge and Electric Flux, p-725
The word “flux” comes from a Latin word meaning “flow.”)
22.3 Three cases in which there is zero net charge inside a box and no
net electric flux through the surface of the box. (a) An empty
box with E=0 (b) A box containing one positive and one equal magnitude
negative point charge. (c) An empty box immersed in a uniform electric
field.
(a) No charge
inside box,
zero flux
(b) Zero net charge inside box,
inward flux cancels outward
flux.
(c) No charge inside box,
inward flux cancels outward
flux.
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3. 22.4 (a) A box enclosing a positive point charge +q (b) Doubling the
charge causes the magnitude of E to double, and it doubles the electric
flux through the surface. (c) If the charge stays the same but the
dimensions of the box are doubled, the flux stays the same. The
magnitude of E on the surface decreases by a factor ¼ of but the area
through which “flows” increases by a factor of 4.
3
(a) A box containing a
charge
(b) Doubling the enclosed
charge doubles the flux.
(c) Doubling the box
dimensions
does not change the
flux.

4. 22.2: Calculating Electric Flux, p-728
Flux: Fluid-Flow Analogy
4
Volume flow rate
Volume flow rate
through A
It is dot product from Fig 25.5 b

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Application:
Flux Through a Basking (stretch out) Shark’s Mouth
Unlike aggressive carnivorous sharks such as great whites, a basking
shark feeds passively on plankton in the water that passes through
the shark’s gills as it swims. To survive on these tiny organisms requires
a huge flux of water through a basking shark’s immense mouth, which
can be up to a meter across. The water flux , the product of the shark’s
speed through the water and the area of its mouth—can be up to 0.5
m3/sec (500 liters per second, or almost 5×105 gallons per hour). In a
similar way, the flux of electric field through a surface depends on the
magnitude of the field and the area of the surface (as well as the
relative orientation of the field and surface).

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Fig. 22.26

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Example
22.1,
p-730

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DATA ?

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Example
22.2

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Gauss’s law is an alternative to Coulomb’s law. While completely
equivalent to Coulomb’s law, Gauss’s law provides a different way
to express the relationship between electric charge and electric
field. It was formulated by Carl Friedrich Gauss (1777–1855), one of
the greatest mathematicians of all time.

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22.10: Carl Friedrich Gauss helped develop several
branches of mathematics, including differential
geometry, real analysis, and number theory. The “bell
curve” of statistics is one of his inventions. Gauss
also made state-of-the-art investigations of the earth’s
magnetism and calculated the orbit of the first asteroid
to be discovered.
Point Charge Inside a Spherical Surface
Gauss’s law states that the total electric flux through any closed
surface (a surface enclosing a definite volume) is proportional to
the total (net) electric charge inside the surface.
In Section 22.1 we observed this relationship qualitatively for
certain special cases; now we’ll develop it more rigorously. We’ll start
with the field of a single positive point charge +q. The field lines
radiate out equally in all directions. We place this charge at the center
of an imaginary spherical surface with radius (R) The magnitude of
the electric field (E) at every point on the surface is given by:

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Here examples from 22.5 to 22.10 (pages 737 to 741) Uni Phy 13th ed., are the
applications of Gauss’s Law. Solve such examples

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Example 22.8
Field between oppositely charged parallel conducting plates
Two large plane parallel conducting plates are given charges of equal
magnitude and opposite sign; the surface charge densities are +σ and –σ
Find the electric field in the region between the plates.

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Example 22.9 Field of a uniformly charged sphere:
Positive electric charge (Q) is distributed uniformly throughout the
volume of an insulating sphere with radius (R) Find the magnitude of
the electric field at a point a distance from the center of the sphere.
22.22 The magnitude of
the electric field of a
uniformly
charged insulating
sphere. Compare this
with the field for a
conducting
sphere (see Fig. 22.18,as
Fig of Example 22.5).

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SOLUTION
IDENTIFY and SET UP: As in Example 22.5, the system is spherically
symmetric. Hence we can use the conclusions of that example about the
direction and magnitude of E To make use of the spherical symmetry, we
choose as our Gaussian surface a sphere with radius r concentric with
the charge distribution.

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Please consult books for detail of each article

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