Chapter 24 Gauss’ Law

A charge inside a box can be probed with a test charge qo to measure E field outside the box.

The volume (V) flow rate (dV/dt) of fluid through the wire rectangle (a) is vA when the area of the rectangle is perpendicular to the velocity vector v and (b) is vA cos φφφφ when the rectangle is tilted at an angle φφφφ.

We will next replace the fluid velocity flow vector v with the electric field vector E to get to the concept of electric flux ΦΦΦΦE.

Volume flow rate through the wire rectangle.

(a) The electric flux through the surface = EA.

(b) When the area vector makes an angle φ with the vector E, the area projected onto a plane oriented perpendicular to the flow is A perp. = A cos φ. The flux is zero when φ = 90o because the rectangle lies in a plane parallel to the flow and no fluid flows through the rectangle

A flat surface in a uniform electric field.

The Porcupine Theory of Gauss’ Law

Imagine that we have some porcupines with very long quills. Careful examination shows each porcupine has exactly 30 quills.

The Porcupine Theory of Gauss’ Law

Imagine that we have some porcupines with very long quills. Careful examination shows each porcupine has exactly 30 quills.

Now suppose that we hide a number of these porcupines in a loose-weave burlap bag. How many porcupines are in the bag?

The Porcupine Theory of Gauss’ Law

Imagine that we have some porcupines with very long quills. Careful examination shows each porcupine has exactly 30 quills.

Now suppose that we hide a number of these porcupines in a loose-weave burlap bag. How many porcupines are in the bag?

Answer: Count the number of quills coming out through the cloth. Then, by dividing the number of quills by 30, we can tell how many porcupines are in the bag.

The Porcupine Theory of Gauss’ LawImagine that we have some porcupines with

very long quills. Careful examination shows each porcupine has exactly 30 quills.

Now suppose that we hide a number of these porcupines in a loose-weave burlap bag. How many porcupines are in the bag?

Answer: Count the number of quills coming out through the cloth. Then, by dividing the number of quills by 30, we can tell how many porcupines are in the bag.

This is a good analogy to Gauss’ Law. The quill-count is the electric flux, the porcupines are electric charges, and the bag is a “Gaussian surface” surrounding the charges. Each charge Q has a “quill count” of Q/εεεε0.

+

+

+

+

The Concept of Electric Flux

The basic idea here is that the qualitative idea of field lines can be quantified by defining electric flux as a conserved quantity that is “delivered” by electric charge. Charge creates flux, and observing flux allows one to deduce the presence of charge.

If an electric field is observed to be coming out of all walls of an opaque box as in (a), we can conclude that the box contains a net positive charge. If an electric field is observed to be going into all walls of an opaque box as in (b), we can conclude that the box contains a net negative charge. If equal amounts of electric field are observed to be entering and exiting the box as in (c), we can conclude that the box contains no net charge.

Conceptual Question 1

The box contains:

(a) A single positive charge; (b) A single negative charge; (c) No charge; (d) A net positive charge; (e) A net negative charge.

Gaussian Surfaces

A “Gaussian surface” is a closed virtual surface that surrounds a region of space that may contain a quantity of charge.

In practice, Gaussian surfaces are useful primarily when the symmetry of the Gaussian surface matches the symmetry of the electric field that passes through the surface and of the charge distribution that produces it, and when the field lines are perpendicular to the surfaces that they cross.

We would like to choose a closed Gaussian surface to assess the flux so that:1.The electric field & surface are ⊥⊥⊥⊥.2.The electric field is constant or zero over each surface element.

Matching the Fieldwith the Gaussian Surface

To do this, we must match the symmetries of the field and the closed surface.

Electric Flux

Remark:

• When an area is constructed such that a closed surface is formed, we shall adopt the convention that the flux lines passing into the interior of the volume are negativeand those passing out of the interior of the volume are positive.

The Electric Flux of a NonuniformElectric Field

For a non-uniform field, the flux canbe calculated by breaking up the area Aof interest into small surface elements dA and integrating the E·dA flux increments over the surface.

e E Aδ δΦ = ⋅rr

e i i ii i

E Aδ δΦ = Φ = ⋅∑ ∑rr

e e

surface surface

d E dAΦ = Φ = ⋅∫ ∫rr

Flux through a Curved Surface

For a non-uniform field plus a curved surface, the same approach works. Break up the area A into small surface elements dA and integrate the E·dA flux increments over the surface.

When E⊥⊥⊥⊥dA the flux contributions are zero.When E||dA the flux contributions are

maximum.

Flux Surface IntegralThe flux is of particular interest when it surrounds a net charge.

Therefore, we will focus on closed Gaussian surfaces. For that case, we write the flux integral as:

Here, the “loop” on the integral sign indicates a surface integral over a closed surface. Since a simple non-intersecting closed surface has a definite inside and outside, we take the vector direction of dAas always pointing toward the outside.

As we will see, this integral is related to the amount of charge enclosed by the surface.

∫ ⋅=Φ AdEelectricnet

rr

The projection of an element of area dA of a sphere of radius R UP onto a concentric sphere of radius 2R. The projection multiplies each linear dimension by 2, so the area element on the larger sphere is 4 dA. The same number of lines of flux pass thru each area element.

Flux ΦΦΦΦE from a point charge q.

Flux through an irregular surface.

The projection of the area element dA onto the

spherical surface is dA cosφ.

Spherical Gaussian surfacesaround (a) positive and (b) negative point charge.

Gauss’ Lawcan be used to calculate the magnitude of the E field vector:

Gauss’ Law

Use the following recipe for Gauss’s Law problems:

1. Carefully draw a figure - location of all charges, direction of electric field vectors E

2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field.

3. Write Gauss Law and perform dot product E ● dA

4. Since you drew the surface in such a way that the magnitude of the Eis constant on the surface, you can factor the |E| out of the integral.

5. Determine the value of Qencl from your figure and insert it into Gauss‘ Law equation.

6. You plug the surface area of your Gaussian surface in for the integral of dA.

Under electrostatic conditions, any excess charge resides entirely on the surface of a solid conductor.

Gaussian surface

The solution of this problem lies in the fact that the electric field inside a conductor is zero and if we place our Gaussian surfaceinside the conductor (where the field is zero), the charge enclosedmust be zero (+ q – q) = 0.

Find electric charge q on surface of hole in the charged conductor.

The E field inside a conducting box (a “Faraday cage”) in an electric field.

A Gaussian surfacedrawn inside the conducting material of which the box is made must have zero electric field on it (field inside a con-ductor is zero). If the Gaussian surfacehas zero field on it, the charge enclosed must be zero per Gauss's Law.

Electric field = zero (electrostatic) inside a solid conducting sphere

Under electrostatic conditions the electric

field inside a solid conducting sphere is

zero. Outside the sphere the electric field drops off as 1 / r2, as though all the excess charge on the sphere were concentrated at

its center.

Consider a conducting spherical shell, which has an inner radius of R1 = 0.07 m, an outer radius of R2 = 0.09 m and contains a charge of Q2 = +18 C. This shell, in turn, encloses a point charge of Q1 = -6 C located at its center, as shown in the diagram to the right.

Using Gauss’ Law, develop an expression which will predict the electric field strength for values of r <R1 and evaluate that expression for a value of r = 0.05 m.

Consider a conducting spherical shell, which has an inner radius of R1 = 0.07 m, an outer radius of R2 = 0.09 m and contains a charge of Q2 = +18 C. This shell, in turn, encloses a point charge of Q1 = -6 C located at its center, as shown in the diagram to the right.

Using Gauss’ Law develop an expression which will predict the electric field strength outside of the spherical shell and evaluate that expression for a value of r = 0.12 m.

On the diagram above sketch the electric field lines everywhere.

What is the electric field strength for R1 < r < R2?

Find the charges on the inner surface and the outer surface of the conducting shell.

Consider a conducting spherical shell, which has an inner radius of R1 = 0.07 m, an outer radius of R2 = 0.09 m and contains a charge of Q2 = +18 C. This shell, in turn, encloses a point charge of Q1 = -6 C located at its center, as shown in the diagram to the right.

On the graph below sketch the electric field strength as a function of distance from the center of the spherical shell.

A coaxial cylindrical Gaussian surfaceis used to find the electric field outside an infinitely long charged

wire.

An infinitely long charged wire has a uniform charge per unit length λ. Find E as a function of r (the distance from the wire)

End view

An Infinite Conducting Plate• Here is a plate (plane) geometry, where the charges are evenly

distributed on a flat surface. The surface charge density is σ

• The E field is everywhere perpendicular to the plate (again, if not, the charges will move until the part parallel to the surface is nullified).

• Use a gaussian surface that is parallel to E on the sides (so no flux through side surfaces), and closes inside the conductor (no flux through that end).

• On the remaining side, the area vector A is parallel to the E field.

Conducting Surface

Electric field between two oppositely charged parallel conducting plates.

Nonconducting SheetA nonconducting sheet with a uniform surface charge density has the same geometry as for the conducting plate, so use the same gaussian surface.

The only difference is that now one end cannot close in a conductor, so there is electric flux out both ends.

As you may expect, the resulting electric field is half of what we got before.

0

2εσA

EAAdE ==⋅∫rr

02εσ=E Nonconducting Sheet of

Charge

Conceptual Question 2How do the

magnitudes of the E-fields compare at the points shown (consider the plate to be infinite in size)?

(a) Eb>Ec>Ed>Ee>Ea;(b) Eb=Ec>Ed=Ee>Ea;(c) Eb=Ec=Ed=Ee>Ea;(d) Eb=Ec=Ed=Ee=Ea;(e) Cannot tell without more information.

Capacitor Edge EffectsSince the electrodes of a real parallel plate capacitor are not infinite, there are “edge

effects” at the ends of the electrodes.

The electric field of a uniformly charged INSULATING sphere.

“Volume charge density”: ρρρρ = charge / unit volume is used to

characterize the charge distribution of a uniformly charged insulator.

Summary for Uniformly Charged ObjectsGauss’ Law states that the electric flux is proportional to the net charge enclosed by the surface, and the constant of proportionality is ε0. In symbols, it is

There are three geometries we typically deal with for charge spread throughout object:

∫ =⋅=Φ00

εεencenc

E

qAdE

q rr

Electric field

Cylindrical, with axis along E.

σ = q/ASheet or Plane

Cylindrical, with axis along line of charge

λ = q/LLinear

Spherical (radius r), with center on center of sphere

ρ = q/VSpherical

Gaussian surfaceCharge DensityGeometry

rE

02πε

λ=

0εσ=E

02εσ=E

2

04 r

qE

πε= r

R

qE

=

3

04πε

Line of Charge

Conducting Nonconducting

r ≥≥≥≥ R r < R