2180 phys lect 3
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Transcript of 2180 phys lect 3
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Gauss’s law and its applications Lect. (3)
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• The product of the magnitude of the electric field E and surface area A perpendicular to the field is called the electric flux E
• the SI units of electric flux (N.m2/C.)• Electric flux is proportional to the
number of electric field• lines penetrating some surface.
Electric Flux
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Electric Flux
• The amount of electric field passing through a surface area• The units of electric flux
are N-m2/C
AE AdE
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Gauss’s Law
The total flux passing through a closed surface is proportional to the charge enclosed within that surface.
Note: The area vector points outward
0q
AdEsurface
E
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The Gaussian Surface
An imaginary closed surface created to enable the application of Gauss’s Law
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1. The Electric Field Due to a Point Charge
2. The Electric Field Due to a Thin Spherical Shell
3. A Cylindrically Symmetric Charge Distribution
Application of Gauss’s Law to Various
Charge Distributions
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1. Spherically Symmetric Charge Distribution
• The area vector is parallel to the electric field vector at the surface of the spherical Gaussian surface.• Applying Gauss’ law:
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3
22in
2in
2in
22
R
rQk
R
rQ
r
k
r
qk
rπ4
qkπ4E
rπ4Eqkπ4
rπ4EAEΦrπ4Asphere1θcos
θcosAEdAEdAEΦ
1.The Electric Field Due to a Point Charge:
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• As r increases to R, the magnitude of the electric field will increase as more charge is included within the volume of the spherical Gaussian surface.• The maximum electric field will be obtained when r = R.• For a spherical Gaussian surface of radius r R, we can consider the charged insulating sphere as a point charge.
1. Spherically Symmetric Charge Distribution
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• Conclusion: the electric field E inside a uniformly charged solid insulating sphere varies linearly with the radius until reaching the surface. The electric field E outside the sphere varies inversely with 1/r2. The diagram on the next slide illustrates this relationship.
22in
2in
2inin
22
r
Qk
r
qk
rπ4
qkπ4
rπ4Eqkπ4qkπ4Φ
rπ4EΦrπ4Asphere1θcos
θcosAEdAEdAEΦ
E
1. Spherically Symmetric Charge Distribution
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Spherically Symmetric Charge Distribution
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2. The Electric Field Due to a Thin Spherical Shell
A thin spherical shell of radius a has a total charge Q distributed uniformly over its surface . Find the electric field at points(A) outside and (B) inside the shell.
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• Consider a thin spherical shell of radius R with a total charge Q distributed uniformly over its surface.
• For points outside the shell, r R: the spherical Gaussian surface will enclose the thin spherical shell, so qin = Q.
2. The Electric Field Due to a Thin Spherical Shell
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• The area vector is parallel to the electric field vector at the surface of the spherical Gaussian surface. The angle between the area vector and the electric field vector is zero.
22in
2in
2inin
22
r
Qk
r
qk
rπ4
qkπ4
rπ4Eqkπ4qkπ4Φ
rπ4EΦrπ4Asphere1θcos
θcosAEdAEdAEΦ
E
2. The Electric Field Due to a Thin Spherical Shell
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• For points inside the thin spherical shell, r < R:
• The charge lies on the surface of the thin spherical shell.
• No charge is found within the shell, so qin = 0 and the electric field and flux within the thin spherical shell is also 0.
2. The Electric Field Due to a Thin Spherical Shell
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Cylindrically Symmetric Charge Distribution
• Consider a uniformly charged cylindrical rod of length L.
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Cylindrically Symmetric Charge Distribution
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Cylindrically Symmetric Charge Distribution
• Instead of the shortened Gaussian cylinder indicated in the diagrams, I use a Gaussian cylinder that is the same length as the charged cylindrical rod.• The electric field lines are perpendicular to the charged
cylindrical rod and are directed outward in all directions. • The electric field is perpendicular to the Gaussian
cylinder and is parallel to the area vector at those points where it passes through the Gaussian cylinder. The angle between the area vector and the electric field vector is zero.
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Cylindrically Symmetric Charge Distribution
• The equation for the area of a cylinder is: A = 2··r·L• Applying Gauss’ law:
r
λk2
Lr
Qk2
Lr
qk2
Lrπ2
qkπ4E
Lrπ2Eqkπ4qkπ4Φ
Lrπ2EΦLrπ2Acylinder1θcos
θcosAEdAEdAEΦ
inin
inin