a b c

Field lines are a way of representing an electric field Lines leave (+) charges and return to (-) charges Number of lines leaving/entering charge  charge Tangent of line = direction of E Local density of field lines  local magnitude of E Lines from an isolated charge go to infinity Field lines cannot cross Last Week Rules for Field Lines + -

Electric Flux Flux: Let’s quantify previous discussion about numbers of field-lines Define: electric flux  s  through the closed surface S “ S ” is surface of the box Note: in Fishbane they use dA to refer to an element of surface area. In these slides I use dS for the same thing

Flux How much of something is passing through some surface Ex: How many hairs passing through your scalp. Two ways to define 1.Number per unit area (e.g., 10 hairs/mm 2 ) This is NOT what we use here. 2.Number passing through an area of interest e.g., 48,788 hairs passing through my scalp. This is what we are using here.

Electric Flux What does this new quantity mean? The integral is over a CLOSED SURFACE Since is a SCALAR product, the electric flux is a SCALAR quantity The integration vector is normal to the surface and points OUT of the surface. is interpreted as the component of E which is NORMAL to the SURFACE

Electric Flux Therefore, the electric flux through a closed surface is the sum of the normal components of the electric field all over the surface. The sign matters!! Pay attention to the direction of the normal component as it penetrates the surface… is it “out of” or “into” the surface? “Out of” is “+” “into” is “-”

How to think about flux Consider the flux through this surface and define the surface area vector Let E-field point in y- direction –then and are parallel and Look at this from on top surface area vector: w w

How to think about flux Consider flux through two surfaces that “intercept different numbers of field lines” –first surface is the surface from the previous slide –Second surface rotated by an angle  case 1 case 2  Case 2 is smaller! case 1 E -field surface area E S Flux: case 2

The Sign Problem For an open surface we can choose the direction of S-vector two different ways –to the left or to the right –what we call flux would be different these two ways –different by a minus sign right left A differential surface element, with its vector For a closed surface we can choose the direction of S -vector two different ways –pointing “in” or “out” –Choose “out” –Integral of E dS over a closed surface gives net flux “out,” but can be + or -

E 1 2 Wire loops (1) and (2) are placed in a uniform electric field as shown. Compare the flux through the two surfaces. a) Ф 1 > Ф 2 b) Ф 1 = Ф 2 c) Ф 1 < Ф 2 Question 1

1.a 2.b 3.c

E 1 2 Wire loops (1) and (2) are placed in a uniform electric field as shown. Compare the flux through the two surfaces. a) Ф 1 > Ф 2 b) Ф 1 = Ф 2 c) Ф 1 < Ф 2 Question 1

A cube is placed in a uniform electric field. Find the flux through the bottom surface of the cube. a) Ф bottom < 0 b) Ф bottom = 0 c) Ф bottom > 0 Question 2

1.a 2.b 3.c

A cube is placed in a uniform electric field. Find the flux through the bottom surface of the cube. a) Ф bottom < 0 b) Ф bottom = 0 c) Ф bottom > 0 Question 2

Question 3 Imagine a cube of side a positioned in a region of constant electric field as shown Which of the following statements about the net electric flux  E through the surface of this cube is true? (a)  E =  (b)  E  2 a  (c)  E  6 a  a a

Question 3 1.a 2.b 3.c

Question 3 Imagine a cube of side a positioned in a region of constant electric field as shown Which of the following statements about the net electric flux  E through the surface of this cube is true? (a)  s =  (b)  s  2 a  (c)  s  6 a  a a The electric flux through the surface is defined by: Therefore, the total flux through the cube is: on the bottom face is negative. ( dS is out; E is in) on the top face is positive. ( dS is out; E is out) is ZERO on the four sides that are parallel to the electric field.

The Fundamental Law of Electrostatics? Coulomb’s Law Force between two point charges OR Gauss’ Law Relationship between Electric Fields and charges

dS 1 2 A positive charge is contained inside a spherical shell. When the charge moves from position 1 to position 2; The electric flux d Ф s through the surface element dS increases BUT the electric flux through the entire surface stays constant

Gauss’ Law Consider the dipole flux passing through the ‘Gaussian’ surfaces –Surface 1 has +ve flux proportional to charge +q –Surface 2 has –ve flux proportional to charge –q –Surface 3 has zero flux, as many flux lines enter as leave –Surface 4 also has zero flux, all leaving lines return Gaussian surface 4

Question 4 R 2R2R Consider 2 spheres (of radius R and 2 R ) drawn around a single charge as shown. –Which of the following statements about the net electric flux through the 2 surfaces (  2 R and  R ) is true? (a)  R <  2 R (b)  R =  2 R (c)  R >  2 R

Question 4 1.a 2.b 3.c

R 2R2R Consider 2 spheres (of radius R and 2 R ) drawn around a single charge as shown. –Which of the following statements about the net electric flux through the 2 surfaces (  2 R and  R ) is true? (a)  R <  2 R (b)  R =  2 R (c)  R >  2 R Question 4 Look at the lines going out through each circle -- each circle has the same number of lines. The electric field is different at the two surfaces, because E is proportional to 1 / r 2, but the surface areas are also different. The surface area of a sphere is proportional to r 2. Since flux =, the r 2 and 1/ r 2 terms will cancel, and the two circles have the same flux! There is an easier way. Gauss’ Law states the net flux is proportional to the NET enclosed charge. The NET charge is the SAME in both cases.

Gauss’ Law Gauss’ Law (a FUNDAMENTAL LAW): The net electric flux through any closed surface is proportional to the charge enclosed by that surface. How do we use this equation?? The above equation is ALWAYS TRUE but it isn’t always easy to use. It is very useful in finding E when the physical situation exhibits strong SYMMETRY.

Gauss’ Law … made easy To solve the above equation for E, you have to be able to CHOOSE A CLOSED SURFACE such that the integral is TRIVIAL. (1) Direction: surface must be chosen such that E is known to be either parallel or perpendicular to each piece of the surface; If then (2) Magnitude: surface must be chosen such that E has the same value at all points on the surface when E is perpendicular to the surface.

With these two conditions we can bring E outside of the integral…and: Note that is just the area of the Gaussian surface over which we are integrating. Gauss’ Law now takes the form: This equation can now be solved for E (at the surface) if we know q enclosed (or for q enclosed if we know E ). Gauss’ Law … made easy

Summary Electric Flux Gauss’ Law Read Chapter 23