a b c Gauss' Law

Today… Gauss’ Law: Motivation & Definition Coulomb’s Law as a consequence of Gauss’ Law Charges on Conductors: Where are they? Applications of Gauss’ Law Uniform Charged Sphere Infinite Line of Charge Infinite Sheet of Charge Two infinite sheets of charge Shortcuts Appendix: Three Gauss’ Laws examples

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

Preflight 4: dS dS 2 1 2) A positive charge is contained inside a spherical shell. How does the electric flux dФE through the surface element dS change when the charge is moved from position 1 to position 2? a) dФE increases b) dФE decreases c) dФE doesn’t change

dS dS 2 1 3) A positive charge is contained inside a spherical shell. How does the flux ФE through the entire surface change when the charge is moved from position 1 to position 2? a) ФE increases b) ФE decreases c) ФE doesn’t change

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 doesn’t look easy to use. It is very useful in finding E when the physical situation exhibits massive 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.

Gauss’ Law…made easy 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 qenclosed (or for qenclosed if we know E).

Geometry and Surface Integrals If E is constant over a surface, and normal to it everywhere, we can take E outside the integral, leaving only a surface area you may use different E’s for different surfaces of your “object” a b c x y z z R R L

Gauss Þ Coulomb We now illustrate this for the field of the point charge and prove that Gauss’ Law implies Coulomb’s Law. Symmetry Þ E-field of point charge is radial and spherically symmetric Draw a sphere of radius R centered on the charge. E +Q R Why? E normal to every point on the surface E has same value at every point on the surface can take E outside of the integral! Therefore, ! Gauss’ Law We are free to choose the surface in such problems… we call this a “Gaussian” surface Þ

Uniform charged sphere What is the magnitude of the electric field due to a solid sphere of radius a with uniform charge density r (C/m3)? Outside sphere: (r>a) We have spherical symmetry centered on the center of the sphere of charge Therefore, choose Gaussian surface = hollow sphere of radius r Þ Gauss’ Law same as point charge!

Uniform charged sphere Outside sphere: (r > a) Inside sphere: (r < a) We still have spherical symmetry centered on the center of the sphere of charge. Therefore, choose Gaussian surface = sphere of radius r Gauss’ Law r a E But, Thus:

Gauss’ Law and Conductors We know that E=0 inside a conductor (otherwise the charges would move). But since  . Charges on a conductor only reside on the surface(s)! Conducting sphere +

Preflight 4: A blue sphere A is contained within a red spherical shell B. There is a charge QA on the blue sphere and charge QB on the red spherical shell. A B 7) The electric field in the region between the spheres is completely independent of QB the charge on the red spherical shell. True False

Lecture 4, ACT 1 Consider the following two topologies: A) A solid non-conducting sphere carries a total charge Q = -3 mC distributed evenly throughout. It is surrounded by an uncharged conducting spherical shell. s2 s1 -|Q| E B) Same as (A) but conducting shell removed 1A Compare the electric field at point X in cases A and B: (a) EA < EB (b) EA = EB (c) EA > EB 1B What is the surface charge density s1 on the inner surface of the conducting shell in case A? (a) s1 < 0 (b) s1 = 0 (c) s1 > 0

Lecture 4, ACT 1 Consider the following two topologies: A solid non-conducting sphere carries a total charge Q = -3 mC distributed evenly throughout. It is surrounded by an uncharged conducting spherical shell. s2 s1 -|Q| E 1A Compare the electric field at point X in cases A and B: (a) EA < EB (b) EA = EB (c) EA > EB Select a sphere passing through the point X as the Gaussian surface. How much charge does it enclose? Answer: -|Q|, whether or not the uncharged shell is present. (The field at point X is determined only by the objects with NET CHARGE.)

Lecture 4, ACT 1 s2 Consider the following two topologies: A solid non-conducting sphere carries a total charge Q = -3 mC and is surrounded by an uncharged conducting spherical shell. B) Same as (A) but conducting shell removed s1 -|Q| E What is the surface charge density s1 on the inner surface of the conducting shell in case A? 1B (a) s1 < 0 (b) s1 = 0 (c) s1 > 0 Inside the conductor, we know the field E = 0 Select a Gaussian surface inside the conductor Since E = 0 on this surface, the total enclosed charge must be 0 Therefore, s1 must be positive, to cancel the charge -|Q| By the way, to calculate the actual value: s1 = -Q / (4 p r12)

Infinite Line of Charge Symmetry Þ E-field must be ^ to line and can only depend on distance from line Therefore, CHOOSE Gaussian surface to be a cylinder of radius r and length h aligned with the x-axis. + + + + + + + + + x y + + + + + + + + + + + Er h + + + + + + Apply Gauss’ Law: On the ends, AND On the barrel, Þ NOTE: we have obtained here the same result as we did last lecture using Coulomb’s Law. The symmetry makes today’s derivation easier.

Lecture 4, ACT 2 A line charge l (C/m) is placed along the axis of an uncharged conducting cylinder of inner radius ri = a, and outer radius ro = b as shown. What is the value of the charge density so (C/m2) on the outer surface of the cylinder? s0=? b a l (a) (b) (c)

Lecture 4, ACT 2 A line charge l (C/m) is placed along the axis of an uncharged conducting cylinder of inner radius ri = a, and outer radius ro = b as shown. What is the value of the charge density so (C/m2) on the outer surface of the cylinder? s0=? b a l (a) (b) (c) View end on: b Draw Gaussian tube which surrounds only the outer edge so

a) a cylinder with its axis along the plane Preflight 4: 5) Given an infinite sheet of charge as shown in the figure. You need to use Gauss' Law to calculate the electric field near the sheet of charge. Which of the following Gaussian surfaces are best suited for this purpose? Note: you may choose more than one answer a) a cylinder with its axis along the plane b) a cylinder with its axis perpendicular to the plane c) a cube d) a sphere

Infinite sheet of charge, surface charge density s Symmetry: direction of E = x-axis Therefore, CHOOSE Gaussian surface to be a cylinder whose axis is aligned with the x-axis. +s A x Conclusion: An infinite plane sheet of charge creates a CONSTANT electric field . Apply Gauss' Law: The charge enclosed = s A Therefore, Gauss’ Law Þ On the barrel, On the ends, E E

Two Infinite Sheets (into the screen) Field outside must be zero. Two ways to see: Superposition Gaussian surface encloses zero charge + s - s E=0 E=0 A Field inside is NOT zero: Superposition Gaussian surface encloses non-zero charge E A

Cylindrical symmetry: Gaussian surface = cylinder of radius r Gauss’ Law: Help for the Homework Problems (See also: mandatory on-line “tutorial exercises”) Gauss’ Law is ALWAYS VALID! What Can You Do With This? If you have (a) spherical, (b) cylindrical, or (c) planar symmetry AND: If you know the charge (RHS), you can calculate the electric field (LHS) If you know the field (LHS, usually because E=0 inside conductor), you can calculate the charge (RHS). LHS: RHS: q = ALL charge inside radius r RHS: q = ALL charge inside radius r, length L RHS: q = ALL charge inside cylinder =s A Spherical Symmetry: Gaussian surface = sphere of radius r Planar Symmetry: Gaussian surface = cylinder of area A Cylindrical symmetry: Gaussian surface = cylinder of radius r

Sheets of Charge σ1 σR σL A B C D Hints: E1 + EL - ER = 0 Uncharged Conductor E1 + EL - ER = 0 σ1 + σ L - σ R = 0 σ L + σ R = 0 (uncharged conductor) σ1 + 2σ L = 0 σ L = σ R = Hints: Assume σ is positive. If it’s negative, the answer will still work. Assume to the right. Use superposition, but keep signs straight Think about which way a test charge would move.

Summary Gauss’ Law: Electric field flux through a closed surface is proportional to the net charge enclosed Gauss’ Law is exact and always true…. Gauss’ Law makes solving for E-field easy when the symmetry is sufficient spherical, cylindrical, planar Gauss’ Law proves that electric fields vanish in conductor extra charges reside on surface

Example 1: spheres A solid conducting sphere is concentric with a thin conducting shell, as shown R1 R2 Q1 Q2 The inner sphere carries a charge Q1, and the spherical shell carries a charge Q2, such that Q2 = -3Q1. How is the charge distributed on the sphere? How is the charge distributed on the spherical shell? What is the electric field at r < R1? Between R1 and R2? At r > R2? What happens when you connect the two spheres with a wire? (What are the charges?) A B C D

A + + + + + + + + How is the charge distributed on the sphere? Q2 Q1 * The electric field inside a conductor is zero. (A) By Gauss’s Law, there can be no net charge inside the conductor, and the charge must reside on the outside surface of the sphere + + + + + + + +

B Q2 How is the charge distributed on the spherical shell? Q1 * The electric field inside the conducting shell is zero. (B) 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. The charges are distributed uniformly over the inner and outer surfaces of the shell, hence and

C Q2 What is the Electric Field at r < R1? Between R1 and R2? At r > R2? * The electric field inside a conductor is zero. (C) r < R1: Inside the conducting sphere (C) Between R1 and R2 : R1 < r < R2 Charge enclosed = Q1 (C) r > R2 Charge enclosed = Q1 + Q2

D What happens when you connect the two spheres with a wire? (What are the charges?) - After electrostatic equilibrium is reached, there is no charge on the inner sphere, and none on the inner surface of the shell. The charge Q1 + Q2 on the outer surface remains. Also, for r < R2 and for r > R2

Let’s try some numbers Q1 = 10mC R1 = 5cm Q2 = -30mC R2 = 7cm sinner = -162 mC/m2 souter = -325 mC/m2 Electric field r < 5cm: Er(r = 4cm) = 0 N/C 5cm < r < 7cm: Er(r = 6cm) = 2.5 x 107 N/C r > 7cm: Er(r = 8cm) = -2.81 x 107 N/C r > 7cm: Er(r = 9cm) = -2.22 x 107 N/C B C D

Example 2: Cylinders sinner souter An infinite line of charge passes directly through the middle of a hollow, charged, CONDUCTING infinite cylindrical shell of radius R. We will 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

souter sinner R l stotal h A How is the charge distributed on the cylindrical shell? What is sinner? What is souter? What is the electric field at r<R? What is the electric field for r>R? B C

souter sinner What is sinner? A1 R l stotal h Qinner = -l h The electric field inside the cylindrical shell is zero. Hence, if we choose as our Gaussian surface a cylinder, which lies inside the cylindrical shell, we know that the net charge enclosed is zero. Therefore, there will be a surface charge density on the inside wall of the cylinder to balance out the charge along the line. The total charge on the enclosed portion (of length h) of the line charge is: Total line charge enclosed = l h Therefore, the charge on the inner surface of the conducting cylindrical shell is Qinner = -l h The total charge is evenly distributed along the inside surface of the cylinder. Therefore, the inner surface charge density sinner is just Qinner divided by total area of the cylinder: sinner = -l h / 2p Rh = -l / 2p R Notice that the result is independent of h.

What is souter? souter sinner stotal A2 R l h We know that the net charge density on the cylinder is stotal. The charge densities on the inner and outer surfaces of the cylindrical shell have to add up to stotal. Therefore, souter =stotal – sinner = stotal +l /(2p R).

What is the Electric Field at r<R? B What is the Electric Field at r<R? h Gaussian surface r l R h Whenever we are dealing with electric fields created by symmetric charged surfaces, we must always first chose an appropriate Gaussian surface. In this case, for r <R, the surface surrounding the line charge is actually a cylinder of radius r. Using Gauss’ Law, the following equation determines the E-field: 2prhEr = qenclosed / eo qenclosed is the charge on the enclosed line charge, which is lh, and (2prh) is the area of the barrel of the Gaussian surface. The result is:

What is the Electric field for r>R? Gaussian surface l R stotal r h As usual, we must first chose a Gaussian surface as indicated above. We also need to know the net charge enclosed in our Gaussian surface. The net charge is a sum of the following: Net charge enclosed on the line: lh Net charge enclosed within Gaussian surface, residing on the cylindrical shell: Q= 2pRh stotal Therefore, net charge enclosed is Q + lh The surface area of the barrel of the Gaussian surface is 2prh Now we can use Gauss’ Law: 2prh E = (Q + lh) / eo You have all you need to find the Electric field now. Solve for Er to find

Let’s try some numbers R = 13 cm h = 168 cm stotal = 528 mC/m2 l = 50mC/m A1 sinner = -61.21 mC/m2 A2 souter = 589.2 mC/m2 B Electric Field (r<R); Er(r = 5cm) = 1.798 x 107 N/C C Electric Field (r>R); Er(r = 20cm) = 4.328 x 107 N/C

Example 3: planes A B C y s 1 s 2 x1 x2 x Suppose there are infinite planes positioned at x1 and x2. The plane at x1 has a positive surface charge density of s1 while the plane at x2 has negative surface charge density of s2. Find: A the x-component of the electric field at a point x>x2 the x-component of the electric field at x1<x<x2 the x-component of the electric field at a point x<x1 B C y s 1 s 2 x1 x2 x

Solutions A Ex(x>x2) y s 1 s 2 E2 E1 x1 x2 x We can use superposition to find . The E-field desired is to the right of both sheets. Therefore; y s 1 s 2 E2 E1 x1 x2 x

B Ex(x1<x<x2) This time the point is located to the left of s2 and to the right of s1, therefore; y s 1 s 2 E2 E1 x1 x2 x

C Ex(x<x1) When the point is located to the left of both sheets; y s 1 s 2 E1 E2 x1 x2 x

Let’s add some numbers... x1= -2m x2= 2m s1 = +2mC/m2 s2 = -3mC/m2 A E1x= 1.130 x 105 N/C Ex= -0.565 x 105 N/C E2x= -1.695 x 105 N/C B E1x= 1.130 x 105 N/C Ex= 2.825 x 105 N/C E2x= 1.695 x 105 N/C C E1x= -1.130 x 105 N/C Ex= 0.565 x 105 N/C E2x= 1.695 x 105 N/C