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**Ch. 24 Electric Flux Gauss's Law**

Electric flux is the amount of electric field going across a surface It is defined in terms of a direction, or normal unit vector, perpendicular to the surface For a constant electric field, and a flat surface, it is easy to calculate Denoted by E Units of Nm2/C When the surface is flat, and the fields are constant, you can just use multiplication to get the flux When the surface is curved, or the fields are not constant, you have to perform an integration Here use A X normal unit vector for A vector of text. Really when everything in integral is constant, you can just multiply. So, if E dot n is constant along surface – you are OK.

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Warmup 03

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**Is the answer independent of the shape?**

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Warmup 03

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Solve on Board

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**Total Flux Out of Various Shapes**

A point charge q is at the “center” of a (a) sphere (b) joined hemispheres (c) cylinder (d) cube. What is the total electric flux out of the shape? b a q q a q

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**Electric Flux (Hard example)**

A point charge q is at the center of a cylinder of radius a and height 2b. What is the electric flux out of (a) each end and (b) the lateral surface? top s b b r r z a Consider a ring of radius s and thickness ds q b a lateral surface

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Gauss’s Law No matter what shape you use, the total electric flux out of a region containing a point charge q is 4keq = q/0. Why is this true? Electric flux is just measuring how many field lines come out of a given region No matter how you distort the shape, the field lines come out somewhere If you have multiple charges inside the region their effects add However, charges outside the region do not contribute q q4 ϕ E = E ∙d A = q in ϵ 0 q3 q1 q2

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Warmup 03

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**Using Gauss’s Law to find total charge**

A cube of side a has an electric field of constant magnitude |E| = E pointing directly out on two opposite faces and directly in on the remaining four faces. What is the total charge inside the cube? A) 6Ea20 B) – 6Ea20 C) 2Ea20 D) – 2Ea20 E) None of the above a a a Do easy problem first – point charge, And problem 4 points on page 696.

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**Using Gauss’s Law to find flux**

A very long box has the shape of a regular pentagonal prism. Inscribed in the box is a sphere of radius R with surface charge density . What is the electric flux out of one lateral side of the box? Perspective view The flux out of the end caps is negligible Because it is a regular pentagon, the flux from the other five sides must be the same End view

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**Using Gauss’s Law to find E-field**

A sphere of radius a has uniform charge density throughout. What is the direction and magnitude of the electric field everywhere? Clearly, all directions are created equal in this problem Certainly the electric field will point away from the sphere at all points The electric field must depend only on the distance Draw a sphere of radius r around this charge Now use Gauss’s Law with this sphere Do point charge first Then do as point charge with q = PV r a Is this the electric field everywhere?

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**Using Gauss’s Law to find E-field (2)**

A sphere of radius a has uniform charge density throughout. What is the direction and magnitude of the electric field everywhere? [Like example 24.3] When computing the flux for a Gaussian surface, only include the electric charges inside the surface r a r/a

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**Electric Field From a Line Charge**

What is the electric field from an infinite line with linear charge density ? L r Electric field must point away from the line charge, and depends only on distance Add a cylindrical Gaussian surface with radius r and length L Use Gauss’s Law The ends of the cylinder don’t contribute On the side, the electric field and the normal are parallel

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**Electric Field From a Plane Charge**

What is the electric field from an infinite plane with surface charge density ? Electric field must point away from the surface, and depends only on distance d from the surface Add a box shaped Gaussian surface of size 2d L W Use Gauss’s Law The sides don’t contribute On the top and bottom, the electric field and the normal are parallel Text uses cylinder. If truly infinite, never far from it. Consider 45 degrees area pf charge increaases as r squared as does 1/E.

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**Text discusses this, page 699**

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Serway 24-35 Solve on Board

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Warmup 04

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**Conductors and Gauss’s Law**

Conductors are materials where charges are free to flow in response to electric forces The charges flow until the electric field is neutralized in the conductor Inside a conductor, E = 0 Draw any Gaussian surface inside the conductor In the interior of a conductor, there is no charge Do all Warmup04 Conductors in equilibrium: (1) E zero inside, (2) charge on surface, (3) outside is sigma/eo, (4) greatest at point The charge all flows to the surface

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Warmup 04

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**Electric Field at Surface of a Conductor**

Because charge accumulates on the surface of a conductor, there can be electric field just outside the conductor Will be perpendicular to surface We can calculate it from Gauss’s Law Draw a small box that slightly penetrates the surface The lateral sides are small and have no flux through them The bottom side is inside the conductor and has no electric field The top side has area A and has flux through it , pg 700 The charge inside the box is due to the surface charge We can use Gauss’s Law to relate these

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**Where does the charge go?**

A hollow conducting sphere of outer radius 2 cm and inner radius 1 cm has q = +80 nC of charge put on it. What is the surface charge density on the inner surface? On the outer surface? A) 20 nC/cm2 B) 5 nC/cm2 C) 4 nC/cm2 D) 0 E) None of the above 80 nC The Gaussian surface is entirely contained in the conductor; therefore E = 0 and electric flux = 0 Therefore, there can’t be any charge on the inner surface 1 cm From the symmetry of the problem, the charge will be uniformly spread over the outer surface 2 cm Point charge q/4Pi eps0 w/ q = 80 pi. Note at r = 2 cm, sigma/eps0 is same: 20/4 cm^2 cutaway view The electric field: The electric field in the cavity and in the conductor is zero The electric field outside the conductor can be found from Gauss’s Law

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Warmup 04

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Serway 24-47 Solve on Board

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Solve on Board Charges set up to neitralize field in conductor.

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a b c 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)

a b c 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)

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