l4 gauss law conductors in electrostatics
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Lecture 4. Conductors in Electrostatics
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General problem of Electrostatics: Given: objects that carry charges (point-like, extended, different materials) Find: the electric field everywhere.
the field equation (Coulombβs Law = Gaussβ Law)
the βmaterialsβ equation (macroscopic description of the matterβs
response to the electric field) + To solve the problem, we need to combine
The simplest materials equation corresponds to the case of metals (conductors) in the electrostatic field.
Iclicker Question
A
B
C
D
E
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Iclicker Question
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For which of the following charge distributions you cannot calculate the electric field using the integral form of Gaussβs law?
A. a uniformly charged sphere of radius R
B. a spherical shell of radius R with charge uniformly distributed over its surface
C. a circular cylinder of radius R and height h with charge uniformly distributed over its surface
D. an infinitely long circular cylinder of radius R with charge uniformly distributed over its surface
E. a uniformly charged plane.
Iclicker Question
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An electric field in a certain region has constant magnitude and direction. From this information alone, we can conclude that
A. there must be positive charge distributed throughout that region
B. there must be negative charge distributed throughout that region
C. there must be zero net charge at all points within that region
D. such an electric field is physically impossible
E. nothing useful about the presence or absence of charge in that region.
Matter in Electrostatic Field
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Mobile electrons Induced/built-in dipoles
Electric field β current Electric field β polarization
Conductors Dielectrics
Though the mechanisms of polarization in metals and dielectrics are different, we can treat metals (for now) as an extreme case of highly-
polarizable dielectrics.
Conductors (Metals) in Electrostatics
+ +
+ +
+ -
- -
-
- E = 0
The equilibrium charge distribution: no net charge in the volume, the surface is charged in such a way that the field of this surface charge compensates the external field in the volume (the electric field lines are terminated and originated at the surface).
Metals: huge number of free-moving electrons (conducting electrons).
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E=0 inside a metal - the βmaterialsβ equation
If a piece of metal is placed in electric field, the electrons will be redistributed (current flows) until the field is 0 everywhere inside the metal.
?
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http://web.mit.edu/viz/EM/visualizations/electrostatics/ChargingByInduction/shielding/shielding.htm
Neutral Conducting Sphere in the Field of a Point Charge
+ Field of Point Charge Field of Surface Charges
Net Field (E=0 inside metal) E=0
Ignore the
cavity!
Surface Charge Density Ο
External field
(without a metal)
π¬ πΈπ =π
2π0
πΈπ =π
2π0
Field of surface charges
π’π’π’π’π’π’: π¬πππ = π¬ βππππ
= π π¨π¨π¨π’π’π’π’: π¬πππ = π¬ +ππππ
=πππ
Superposition:
Metal π
π’π’π’π’π’π’ π¨π¨π¨π’π’π’π’
π¬ π
π
π
π = ππππ¬ π¬πππ = ππ¬
π¬πππ
Comparison: charged metallic and dielectric spheres
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Q
Q Q
Q
Q r
Q
Charged conducting sphere Uniformly charged (dielectric!) sphere
The same field outside if the net Q is the same.
Q
How to Draw Field Lines near Conducting Surfaces
At metallic surfaces the electric field is directed along the normal vector to the surface.
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Metal
π¬πππ ?
Metal
π¬πππ would drive surface current
Metal
π¬πππ the only option
The surface charge density (and, thus, the field intensity) is higher near protrusions: the electrons arrange themselves in the way that minimizes the total energy of this charge distribution.
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http://web.mit.edu/viz/EM/visualizations/electrostatics/ChargingByInduction/shielding/shielding.htm
Neutral Conducting Shell in the Field of a Point Charge
+ Field of Point Charge Field of Surface Charges
Net Field: E=0 everywhere inside E=0
Faraday Cage (Shield)
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Man-made βlightningββ¦
.. and a real one
Faraday cage in my lab
Cavities (no Charges within the Cavity)
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This conclusion holds for any inner surface: the charge density on the inner surface is zero provided no charges within the cavity.
Shell charge: +Q
π π = π
π π = π
= 0
=+π
4ππ2
π
π
ππππ = +π = πππ + ππππ
Charges within Cavities
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+Q + +
+ +
+ -
- -
-
- E=0
In general, it is difficult to calculate the charge distribution on the surface of a conductor placed in an external electric field (see the Figure).
-
+ -
+
+ E=0
Gaussian surface
-
+Q
We can easily find the net surface charge if we place a charge inside a metallic shell. Our logic:
E=0 inside metal
Ξ¦πΈ = 0 for any Gaussian surface inside metal
πππππ = 0 = +π + πππ πππ = βπ
ππππ = βπππ = +π ππππππ = 0 = πππ + ππππ
πππ = βπ
ππππ = +π
Metal Shell as a βBlack Holeβ
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Οout does not depend on the position of Q inside - for a spherical shell Οout must be uniform (this minimizes the energy of this system of charges).
The field outside is completely insensitive to the position/motion of charge Q inside!
Thus, the metal shell divides the whole space into two regions: - βOutsideβ knows about the total charge enclosed, but cannot detect the
charge position - βInsideβ knows nothing about the magnitude and positions of the outside
charges.
Iclicker question
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E=0 -
- - -
- -
- -
an uncharged
+
+
+ +
+
+
+ + Q=0
Problem 22.46: Concentric Spherical Shells
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πΈ π < π = 0
π π = π
π π = π
πΈ π < π < π =π
4ππ0π2
πΈ π > π = 0
πΈ π = π+ =
π π = ππ0
=π
4ππ0π2
14ππ0
ππ2οΏ½π=π
=π
4ππ0π2
Charge densities:
= 0
=+π
4ππ2
π π = π =βπ
4ππ2
π π = π = 0
Electric fields:
π¬ π
π π π π π
Inner shell: +Q
Outer shell: -Q
Conclusion
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Next time: Lecture 5. Electric Potential §§ 23.1-23.4
Metals in electrostatic fields: E = 0 inside.
Appendix I. More Complex Cases Field calculations based on the integral form of GL are not limited to these trivial cases: any charge distribution that can be considered as superposition of symmetrical charge distributions can be treated on the basis of this law.
Ο
2Ο
1. Two infinite planes with the charge
densities Ο and 2Ο.
πΈ =π 1 + 4
2π0
2. Uniformly charged spherical shell with a small hole in it. Find E in the hole. Hole can be considered as (almost flat) βpatchβ of negative charge density (- Ο) on top of a positively charged (Ο) whole sphere.
Field of the sphere πΈπ = 0
πΈπ =π
4ππ0π 2=ππ0
πΈπ =π
2π0
πΈπ =π
2π0
Field of the patch
Superposition! πΈπππ =π
2π0
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Appendix II: Electric Field of a Charged Sphere with a Spherical Cavity
Consider a uniformly charged sphere with a spherical cavity. Calculate the electric field intensity along the dashed line.
Hint: treat cavity as a superposition of positively and negatively charged spheres.
R
E(x)
π4ππ0π
πΈ π₯ π₯>π =π
4ππ0π₯2β
π/84ππ0 π₯ β π /2 2
R
R/2
π4ππ0π
βπ/8
4ππ0π /2
positively charged sphere without a cavity
x
negatively charged cavity
sphere + cavity
βπ/8
4ππ0π /2
R/2
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Appendix III. Conductors (Metals) in Electrostatics
+ +
+ +
+ -
- -
-
- E=0
Letβs consider an uncharged piece of metal (qnet=0), and place this metal in an external electric field. The mobile electrons (negative charges) will move with respect to the positively charged ions (polarization)! The equilibrium charge distribution: no net charge in the volume, the surface is charged in such a way that the field of this surface charge compensates the external field in the volume (the electric field lines are terminated and originated at the surface).
t Letβs consider external E=106N/C=106V/m= 10kV/cm
π = πΈπ0 = 106ππ Γ 9 β 10β12
πΆπ β π2 β 10β5
πΆπ2 = 10
ππΆπ2
π‘ =π
π β πππππππ‘π. =10β5 πΆ
π2
1.6 β 10β19 πΆ Γ 1 β 1028 π3 β 10β14π - the size of a nucleus!
Metals: huge number of free-moving electrons (conducting electrons), ~ 1028 electrons/m3.
High density of mobile charges ensures that any external electric field, no matter how strong, will be Β«screenedΒ» by a very thin surface charge sheet:
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E=0 inside a metal βmaterialsβ boundary condition
Appendix IV. Parallel-Plate Capacitor
Parallel-plate capacitor: (two metallic plates with charges +q and βq)
Gaussian surfaces
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S1: charge density Ο, inside E=0, outside πΈ = ππ0π₯οΏ½
+q βq
S2: charge density 0, inside E=0, outside E=0
S3: charge density 0, inside E=0, outside E=0
S4: charge density - Ο, inside E=0, outside πΈ = ππ0π₯οΏ½