Gauss

USING GAUSS' LAW for ELECTRIC FIELDS

What is Gauss' Law?

GAUSS' LAW states that the electric flux through a closed surface depends upon the charge enclosed in that surface.

When is it used?

GAUSS' LAW will always work but is best used when there is geometric symmetry available around the distribution of charge.

Why is it useful?

GAUSS' LAW simplifies the calculations for many E Fields especially those most commonly found in electronics!

WHAT IS A GAUSSIAN SURFACE?

It is a closed surface with a defined inside and a defined outside.
For ease of calculation and symmetry, 3 Gaussian surfaces are most commonly used. They are as follows: spherical, cylindrical, and rectangular solid.
Such a surface must enclose charge inside itself to measure an ELECTRIC FIELD.
If it does not enclose charge then it measures a field from
the enclosed charge as zero!
A Gaussian Surface is three dimensional!
It is imaginary & is created to measure Electric Fields

WHAT IS AN AREA VECTOR?

An Area Vector has magnitude and direction! It is a vector!

The magnitude of the area vector A is equal to the value
of the area of the surface. Common areas of Gaussian surface are

1) A = 4 p R² for a Sphere.

2) A (caps) = p R² and A = 2 p R L (lateral surface area) for a Cylinder.

3) A = sum of base x height for each side (Rectangular Solid)

The area vector has a direction that points from inside to outside of the Gaussian surface and is ALWAYS perpendicular to the surface!

Area Vector A

The  AREA VECTOR for the top, dark blue face of the cube shown has magnitude of (1.4 m) * (1.4m) = 1.96 m ² with a direction parallel to the z axis (or 1.96 k m ²)

ELECTRIC FLUX

Electric Flux is symbolized by the Greek letter phi or f.

Electric Flux is the dot product of the Electric Field Vector E and the
Area Vector A f = E "dot" A

f depends only on q enclosed/ eo

In this diagram, the Electric Field, E = 5 i, is in the + x direction and has magnitude of 5 N/C.

The only FLUX or flow of the E Field is through the left and right faces on the box.

The Area Vectors are - 4.0 m^2 î (left face) and + 4.0 m^2 î through the right face.

The flux through the left face is equal to 5 * 4 cos 180 = -20 Nm^2/C.

The flux through the right face is 5 * 4 cos 0 = +20 NM^2/C.

The net flux is zero for this uniform field through this Gaussian Surface.

.

For a spherical surface, the dA vector is along the radius to the sphere. As you can see for the point charge + Q , the Area Vector DA and the E field vector are parallel and so the angle between them is zero. This always happens for a point charge or for uniform spherical charge distribution denoted by either sigma s * or rho r*.
*See chart below.

The net flux through any Gaussian surface can also be calculated by dividing the charge enclosed in the Gaussian surface by epsilon subzero or the permitivity of free space (8.85 x 10 ^-12 Coulomb squared per Newton-square meter).

In this case --unlike the product of the Electric Field and the Area Vector-- the distribution of the charge does not require symmetry to provide ease of calculation for Electric Flux.

The electric flux for the above surface is Q3 / epsilon zero.

STEPS FOR A PROOF WITH GAUSS' LAW

The following is the Gaussian proof of the Electric Field for a positive point charge.

PROTOCOL

STEPS

1. Start with a statement of Gauss's law.

2. Perform the dot product with E and DA Usually the angle is 0, 90 or 180 degrees between the E and DA vectors.

3. Pull the E vector to the front of the integral as it should be constant for the Gaussian surface you chose.

4. Evaluate the surface integral based on the geometry of the Gaussian surface you chose. Rewrite "q enclosed" if necessary.

5. Solve for E, the Electric Field.

NOTE THE FOLLOWING CHARGE DISTRIBUTIONS FOR OTHER PROOFS!

LINKS:

A TUTORIAL FOR CHAPTER 24 in Fundamentals of Physics

CONDUCTORS & GAUSS' LAW

A REVIEW OF DOT PRODUCTS

ADVANCED THEORY (& SONGS TO MATCH!)

GREAT REVIEW SITE-- GOOD FOR TEST OVERVIEW

GAUSS' LAW WORKSHOP

STEPS FOR A PROOF WITH GAUSS' LAW The following is the Gaussian proof of the Electric Field for a positive point charge.
PROTOCOL	STEPS
1. Start with a statement of Gauss's law.
2. Perform the dot product with E and DA Usually the angle is 0, 90 or 180 degrees between the E and DA vectors.
3. Pull the E vector to the front of the integral as it should be constant for the Gaussian surface you chose.
4. Evaluate the surface integral based on the geometry of the Gaussian surface you chose. Rewrite "q enclosed" if necessary.
5. Solve for E, the Electric Field.
NOTE THE FOLLOWING CHARGE DISTRIBUTIONS FOR OTHER PROOFS!

USING GAUSS' LAW for ELECTRIC FIELDS

What is Gauss' Law?

GAUSS' LAW states that the electric flux through a closed surface depends upon the charge enclosed in that surface.

When is it used?

GAUSS' LAW will always work but is best used when there is geometric symmetry available around the distribution of charge.

Why is it useful?

GAUSS' LAW simplifies the calculations for many E Fields especially those most commonly found in electronics!

WHAT IS A GAUSSIAN SURFACE?

It is a closed surface with a defined inside and a defined outside.

For ease of calculation and symmetry, 3 Gaussian surfaces are most commonly used. They are as follows: spherical, cylindrical, and rectangular solid.

Such a surface must enclose charge inside itself to measure an ELECTRIC FIELD. If it does not enclose charge then it measures a field from the enclosed charge as zero!

A Gaussian Surface is three dimensional!

It is imaginary & is created to measure Electric Fields

WHAT IS AN AREA VECTOR?

An Area Vector has magnitude and direction! It is a vector!

The magnitude of the area vector A is equal to the value of the area of the surface. Common areas of Gaussian surface are

1) A = 4 p R2 for a Sphere.

2) A (caps) = p R2 and A = 2 p R L (lateral surface area) for a Cylinder.

3) A = sum of base x height for each side (Rectangular Solid)

The area vector has a direction that points from inside to outside of the Gaussian surface and is ALWAYS perpendicular to the surface!

Area Vector A

The AREA VECTOR for the top, dark blue face of the cube shown has magnitude of (1.4 m) * (1.4m) = 1.96 m 2 with a direction parallel to the z axis (or 1.96 k m 2)

ELECTRIC FLUX

Electric Flux is symbolized by the Greek letter phi or f.

Electric Flux is the dot product of the Electric Field Vector E and the Area Vector A f = E "dot" A

f depends only on q enclosed/ eo

In this diagram, the Electric Field, E = 5 i, is in the + x direction and has magnitude of 5 N/C.

The only FLUX or flow of the E Field is through the left and right faces on the box.

The Area Vectors are - 4.0 m^2 î (left face) and + 4.0 m^2 î through the right face.

The flux through the left face is equal to 5 * 4 cos 180 = -20 Nm^2/C.

The flux through the right face is 5 * 4 cos 0 = +20 NM^2/C.

The net flux is zero for this uniform field through this Gaussian Surface.

.

The net flux through any Gaussian surface can also be calculated by dividing the charge enclosed in the Gaussian surface by epsilon subzero or the permitivity of free space (8.85 x 10 ^-12 Coulomb squared per Newton-square meter).

In this case --unlike the product of the Electric Field and the Area Vector-- the distribution of the charge does not require symmetry to provide ease of calculation for Electric Flux.

The electric flux for the above surface is Q3 / epsilon zero.

STEPS FOR A PROOF WITH GAUSS' LAW

The following is the Gaussian proof of the Electric Field for a positive point charge.

PROTOCOL

STEPS

1. Start with a statement of Gauss's law.

2. Perform the dot product with E and DA Usually the angle is 0, 90 or 180 degrees between the E and DA vectors.

3. Pull the E vector to the front of the integral as it should be constant for the Gaussian surface you chose.

4. Evaluate the surface integral based on the geometry of the Gaussian surface you chose. Rewrite "q enclosed" if necessary.

5. Solve for E, the Electric Field.

NOTE THE FOLLOWING CHARGE DISTRIBUTIONS FOR OTHER PROOFS!

LINKS:

A TUTORIAL FOR CHAPTER 24 in Fundamentals of Physics

CONDUCTORS & GAUSS' LAW

A REVIEW OF DOT PRODUCTS

ADVANCED THEORY (& SONGS TO MATCH!)

GREAT REVIEW SITE-- GOOD FOR TEST OVERVIEW

GAUSS' LAW WORKSHOP

Such a surface must enclose charge inside itself to measure an ELECTRIC FIELD.
If it does not enclose charge then it measures a field from
the enclosed charge as zero!

The magnitude of the area vector A is equal to the value
of the area of the surface. Common areas of Gaussian surface are

1) A = 4 p R² for a Sphere.

2) A (caps) = p R² and A = 2 p R L (lateral surface area) for a Cylinder.

The AREA VECTOR for the top, dark blue face of the cube shown has magnitude of (1.4 m) * (1.4m) = 1.96 m ² with a direction parallel to the z axis (or 1.96 k m ²)

Electric Flux is the dot product of the Electric Field Vector E and the
Area Vector A f = E "dot" A