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 (Coulomb’s Law = Gauss’ Law)

the “materials” equation (macroscopic description of the matter’s

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 Gauss’s 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 → current Electric field → 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 “materials” 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 σ

External field

(without a metal)

𝑬 𝐸𝑆 =𝜎

2𝜖0

𝐸𝑆 =𝜎

2𝜖0

Field of surface charges

𝐢𝐢𝐢𝐢𝐢𝐢: 𝑬𝒏𝒏𝒏 = 𝑬 −𝝈𝟐𝝐𝟎

= 𝟎 𝐨𝐨𝐨𝐢𝐢𝐢𝐢: 𝑬𝒏𝒏𝒏 = 𝑬 +𝝈𝟐𝝐𝟎

=𝝈𝝐𝟎

Superposition:

Metal 𝒚

𝐢𝐢𝐢𝐢𝐢𝐢 𝐨𝐨𝐨𝐢𝐢𝐢𝐢

𝑬 𝒚

𝒚

𝝈

𝝈 = 𝟐𝝐𝟎𝑬 𝑬𝒏𝒏𝒏 = 𝟐𝑬

𝑬𝒏𝒏𝒏

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

𝑬𝒏𝒏𝒏 ?

Metal

𝑬𝒏𝒏𝒏 would drive surface current

Metal

𝑬𝒏𝒏𝒏 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 “lightning”…

.. 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

𝜎 𝑟 = 𝑎

𝜎 𝑟 = 𝑏

= 0

=+𝑄

4𝜋𝑏2

𝑎

𝑏

𝑄𝑛𝑛𝑛 = +𝑄 = 𝑄𝑖𝑛 + 𝑄𝑜𝑜𝑛

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

Φ𝐸 = 0 for any Gaussian surface inside metal

𝑄𝑛𝑛𝑒𝑒 = 0 = +𝑄 + 𝑄𝑖𝑛 𝑄𝑖𝑛 = −𝑄

𝑄𝑜𝑜𝑛 = −𝑄𝑖𝑛 = +𝑄 𝑄𝑚𝑛𝑛𝑚𝑒 = 0 = 𝑄𝑖𝑛 + 𝑄𝑜𝑜𝑛

𝑄𝑖𝑛 = −𝑄

𝑄𝑜𝑜𝑛 = +𝑄

Metal Shell as a “Black Hole”

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σout does not depend on the position of Q inside - for a spherical shell σout 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: - “Outside” knows about the total charge enclosed, but cannot detect the

charge position - “Inside” 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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𝐸 𝑟 < 𝑏 = 0

𝜎 𝑟 = 𝑎

𝜎 𝑟 = 𝑏

𝐸 𝑏 < 𝑟 < 𝑐 =𝑄

4𝜋𝜀0𝑟2

𝐸 𝑟 > 𝑐 = 0

𝐸 𝑟 = 𝑏+ =

𝜎 𝑟 = 𝑏𝜀0

=𝑄

4𝜋𝜀0𝑏2

14𝜋𝜀0

𝑄𝑟2�𝑟=𝑏

=𝑄

4𝜋𝜀0𝑏2

Charge densities:

= 0

=+𝑄

4𝜋𝑏2

𝜎 𝑟 = 𝑐 =−𝑄

4𝜋𝑐2

𝜎 𝑟 = 𝑑 = 0

Electric fields:

𝑬 𝒓

𝒓 𝑎 𝑏 𝑐 𝑑

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.

σ

2σ

1. Two infinite planes with the charge

densities σ and 2σ.

𝐸 =𝜎 1 + 4

2𝜀0

2. Uniformly charged spherical shell with a small hole in it. Find E in the hole. Hole can be considered as (almost flat) “patch” of negative charge density (- σ) on top of a positively charged (σ) whole sphere.

Field of the sphere 𝐸𝑠 = 0

𝐸𝑠 =𝑄

4𝜋𝜖0𝑅2=𝜎𝜖0

𝐸𝑝 =𝜎

2𝜖0

𝐸𝑝 =𝜎

2𝜖0

Field of the patch

Superposition! 𝐸𝑛𝑛𝑛 =𝜎

2𝜖0

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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)

𝑄4𝜋𝜖0𝑅

𝐸 𝑥 𝑥>𝑅 =𝑄

4𝜋𝜖0𝑥2−

𝑄/84𝜋𝜖0 𝑥 − 𝑅/2 2

R

R/2

𝑄4𝜋𝜖0𝑅

−𝑄/8

4𝜋𝜖0𝑅/2

positively charged sphere without a cavity

x

negatively charged cavity

sphere + cavity

−𝑄/8

4𝜋𝜖0𝑅/2

R/2

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Appendix III. Conductors (Metals) in Electrostatics

+ +

+ +

+ -

- -

-

- E=0

Let’s 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 Let’s consider external E=106N/C=106V/m= 10kV/cm

𝜎 = 𝐸𝜖0 = 106𝑉𝑚 × 9 ∙ 10−12

𝐶𝑁 ∙ 𝑚2 ≈ 10−5

𝐶𝑚2 = 10

𝜇𝐶𝑚2

𝑡 =𝜎

𝑒 ∙ 𝑐𝑐𝑐𝑐𝑒𝑐𝑡𝑟. =10−5 𝐶

𝑚2

1.6 ∙ 10−19 𝐶 × 1 ∙ 1028 𝑚3 ≈ 10−14𝑚 - 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 «screened» by a very thin surface charge sheet:

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E=0 inside a metal “materials” boundary condition

Appendix IV. Parallel-Plate Capacitor

Parallel-plate capacitor: (two metallic plates with charges +q and –q)

Gaussian surfaces

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S1: charge density σ, inside E=0, outside 𝐸 = 𝜎𝜖0𝑥�

+q –q

S2: charge density 0, inside E=0, outside E=0

S3: charge density 0, inside E=0, outside E=0

S4: charge density - σ, inside E=0, outside 𝐸 = 𝜎𝜖0𝑥�