Gauss’s law : introduction Chapter 21: Gauss’s Law Gauss’s law : introduction The total electric flux through a closed surface is equal to the total (net) electric charge inside the surface divided by e0 Gauss’s law is equivalent to Coulomb’s law

Gauss’s law : introduction (cont’d) Consider a distribution of charge Surround it with an imaginary surface that encloses the charge Look at the electric field at various points on this imaginary surface To find out charge distribution inside the imaginary surface, we need to measure electric fields especially on the surface q To do that place a test charge of a known charge amount and measure the electric force + imaginary surface + test charge

Charge and Electric Flux Electric fields by different charges outward flux outward flux + + + inward flux - - inward flux -

Charge and Electric Flux outward flux outward flux + + + When the distance to the surface doubled, the area of the surface quadrupled and the electric field becomes ¼. +

Charge and Electric Flux Definition of electric flux For any point on a small area of a surface, take the product of the average component of E perpendicular to the surface and the area. Then the sum of this quantity over the surface is the net electric flux A qualitative statement of Gauss’s law Whether there is a net outward or inward electric flux through a closed surface depends on the sign of the enclosed charge. Charges outside the surface do not give a net electric flux through the surface. The net electric flux is directly proportional to the net amount of charge enclosed within the surface but is otherwise independent of the size of the closed surface.

Calculating Electric Flux Analogy between electric flux and field of velocity vector A good analogy between the electric flux and the field of velocity vector in a flowing fluid can be found. A (area) volume flow rate: velocity vector (flow speed) A vector area that defines the plane of the area, perpendicular to the plane volume flow rate: f

Calculating Electric Flux Analogy between electric flux and field of velocity vector volume flow rate: A (area) electric flux: Electric field vector A vector area that defines the plane of the area, perpendicular to the plane E field is perpendicular to this plane electric flux: f

Calculating Electric Flux A small area element and flux Total flux for an area Example: Electric flux through a disk r = 0.10 m

Calculating Electric Flux Example : Electric flux through a cube L

Calculating Electric Flux Example : Electric flux through a sphere r=0.20 m + +q q=3.0 mC A=4pr2

Gauss’s Law Preview: The total electric flux through any closed surface (a surface enclosing a definite volume) is proportional to the total (net) electric charge inside the surface. Case 1: Field of a single positive charge q A sphere with r=R at r=R r=R + The flux is independent of the radius R of the surface.

Gauss’s Law Every field line that passes through Case 1: Field of a single positive charge q (cont’d) Every field line that passes through the smaller sphere also passes through the larger sphere The total flux through each sphere is the same r=R The same is true for any portion of its surface such as dA + r=2R This is true for a surface of any shape or size provided it is a closed surface enclosing the charge

Gauss’s Law Case 2: Field of a single positive charge (general surface) + surface perpendicular to +

Gauss’s Law Electric field lines that go in come out. Case 3: An closed surface without any charge inside Electric field lines that go in come out. Electric field lines can begin or end inside a region of space only when there is charge in that region. + Gauss’s law The total electric flux through a closed surface is equal to the total (net) electric charge inside the surface divided by e0

Applications of Gauss’s Law Introduction The charge distribution the field The symmetry can simplify the procedure of application Electric field by a charge distribution on a conductor When excess charge is placed on a solid conductor and is at rest, it resides entirely on the surface, not in the interior of the material (excess charge = charge other than the ions and free electrons that make up the material conductor A Gaussian surface inside conductor Charges on surface Conductor

Applications of Gauss’s Law Electric field by a charge distribution on a conductor (cont’d) A Gaussian surface inside conductor Charges on surface Conductor E at every point in the interior of a conducting material is zero in an electrostatic situation (all charges are at rest). If E were non-zero, then the charges would move Draw a Gaussian surface inside of the conductor E=0 everywhere on this surface (inside conductor) The net charge inside the surface is zero There can be no excess charge at any point within a solid conductor Any excess charge must reside on the conductor’s surface E on the surface is perpendicular to the surface Gauss’s law

Applications of Gauss’s Law Example: Field of a charged conducting sphere with q Gaussian surface E + + + Draw a Gaussian surface outside the sphere + + R + + + r R 2R 3R

Applications of Gauss’s Law Example: Field of a line charge Gaussian surface chosen according to symmetry line charge density

Applications of Gauss’s Law Example: Field of an infinite plane sheet of charge + + + + + + area A + area A two end surfaces + + + + Gaussian surface

Applications of Gauss’s Law Example : Field between oppositely charged parallel conducting plane plate 1 plate 2 Solution 1: - + b - + a - c + outward flux - + S1 S4 - + - + - S2 + S3 - + inward flux - + Solution 2: At Point a : No electric flux on these surfaces b : c :

Applications of Gauss’s Law Example : Field of a uniformly charged sphere Gaussian surface + + + + + + + + r=R + + + + R

Applications of Gauss’s Law Example : Field of a hollow charged (uniformly on its surface) sphere r=0.300 m R=0.250 m Gaussian surface Hollow charged sphere

Charges on Conductors its outer surface in an electrostatic situation Case 1: charge on a solid conductor resides entirely on its outer surface in an electrostatic situation + + + + The electric field at every point within a conductor is zero and any excess charge on a solid conductor is located entirely on its surface. + + + + + + + + + + + + Case 2: charge on a conductor with a cavity + + + + + + If there is no charge within the cavity, the net charge on the surface of the cavity is zero. + + + + + + + + + + Gauss surface

Charges on Conductors inside the cavity Case 3: charge on a conductor with a cavity and a charge q inside the cavity The conductor is uncharged and insulated from charge q. + + + + + + - + - The total charge inside the Gauss surface should be zero from Gauss’s law and E=0 on this surface. Therefore there must be a charge –q distributed on the surface of the cavity. - + - + - + - - + + + + + + + Gauss surface The similar argument can be used for the case where the conductor originally had a charge qC. In this case the total charge on the outer surface must be q+qC after charge q is inserted in cavity.

Charges on Conductors charged conducting ball conductor Faraday’s ice pail experiment charged conducting ball conductor (1) Faraday started with a neutral metal ice pail (metal bucket) and an uncharged electroscope. (2) He then suspended a positively charged metal ball into the ice pail, being careful not to touch the sides of the pail. The leaves of the electroscope diverged. Moreover, their degree of divergence was independent of the metal ball's exact location. Only when the metal ball was completely withdrawn did the leaves collapse back to their original position.

Charges on Conductors charged conducting ball conductor Faraday’s ice pail experiment (cont’d) charged conducting ball conductor (3) Faraday noticed that if the metal ball was allowed to contact the inside surface of the ice pail, the leaves of the electroscope remained diverged (4) Afterwards, when he completely removed the ball from the inside of the ice pail, the leaves remained diverged. However, the metal ball was no longer charged. Since the leaves of the electroscope that was attached to the OUTSIDE of the pail did not move when the ball touched the inside of the pail, he concluded that the inner surface had just enough charge to neutralize the ball.

Charges on Conductors The electric field just outside a conductor has Field at the surface of a conductor The electric field just outside a conductor has magnitude  /0 and is directed perpendicular to the surface. Draw a small pill box that extends into the conductor. Since there is no field inside, all the flux comes out through the top. EA=q/0= A/ 0, E=  / 0

Exercises Exercise 1 This is the same as the field due to a point charge with charge +2Q

Exercises Exercise 1 (cont’d)

Exercises Exercise 2: A sphere and a shell of conductor Q2=-3Q1 From Gauss’s law there can be no net charge inside the conductor, and the charge must reside on the outside surface of the sphere Q2 Q1 R1 There can be no net charge inside the conductor. Therefore the inner surface of the shell must carry a net charge of –Q1 , and the outer surface must carry the charge +Q1+Q2 so that the net charge on the shell equals Q2 . These charges are uniformly distributed. R2

Exercises Exercise 2: A sphere and a shell of conductor (cont’d) Q2=-3Q1 Q2 Q1 R1 R2

Exercises sinner souter stotal R l h Exercise 3: Cylinder An infinite line of charge passes directly through the middle of a hallow charged conducting infinite cylindrical shell of radius R. Let’s focus on a segment of the cylindrical shell of length h. The line charge has a linear charge density l, and the cylindrical shell has a net surface charge density of stotal. stotal R l sinner souter h

Exercises sinner souter stotal R l h Exercise 3: Cylinder (cont’d) The electric field inside the cylindrical shell is zero. Therefore if we choose as a Gaussian surface a cylinder, which lies inside the cylindrical shell, the net charge enclosed is zero. There is a surface charge density on the inside wall of the cylinder to balance out the charge along the line. stotal R l sinner souter h

Exercises sinner souter stotal R l h Exercise 3: Cylinder (cont’d) The total charge on the enclosed portion (length h) of the line charge: The charge on the inner surface of the conducting cylinder shell: stotal R l sinner souter h

Exercises sinner souter stotal R l h Exercise 3: Cylinder (cont’d) The net charge density on the cylinder: The outer charge density : stotal R l sinner souter h

Exercises sinner souter stotal R l h Exercise 3: Cylinder (cont’d) Draw a Gaussian surface surrounding the line charge of radius r (< R) stotal R l sinner souter h

Exercises sinner souter stotal R l h Exercise 3: Cylinder (cont’d) Draw a Gaussian surface surrounding the line charge of radius r (>R) Net charge enclosed on the line: Net charge enclosed on the shell: stotal R l sinner souter h