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Unformatted text preview: Gauss’ Gauss’ Law.
Deutsche Mark 19482002 Johann Carl Friedrich Gauss 17771855 Gauss’ Law: 1835  present 1835
If you are planning on taking Phys 214, 237 etc. you will learn that there are charges other than electric. For each charge there is a version of Gauss’ Law. Gauss’ Law occurs in e.g. (Quantum) Field Theory and Quantum Gravity.
1 Last Last time.
We learned how to compute electric field created by an arbitrary charge distribution. In general, need to compute complicated integrals. Today. Today.
We will NOT compute complicated integrals. In fact, we will hardly do any involved calculation at all. We WILL draw lots of pictures. 2 Notion Notion of Flux.
Define “human flux” as the total number of “Nittany Lions” fans entering the Beaver Stadium per minute. What does this flux depend on? number of entrances For instance, doubling the number of entrances would double the flux. Flux number of fan lines
3 Electric Electric Flux. Definition. Definition.
Uniform Field
A E E A n it is convenient to introduce: A=An, n = unit normal θ=angle between E & n Flux: Ф:=EA
S Flux: Ф:=E·A=EAcosθ NonUniform Field
Idea:
1. 2. 3. Break the surface into small patches Compute elementary fluxes dФ=E·dA Sum all the elementary contributions E Remark: there is an ambiguity in the direction of the normal there
flux is determined up to a sign
4 Electric Electric Flux. Closed surface. Closed By convention, the normal is taken to be outward. In general, the flux integral is not easy to compute, unless there is symmetry. Sample problem:
Electric field is induced by a point charge Q. Compute electric flux through a sphere of radius R centered at the charge. 5 Electric Electric Flux. Geometrical meaning. Geometrical
Uniform Field
A E E 2A 2E A Ф0=EA Ф=2Ф0 Ф=2Ф0 Flux number of field lines crossing surface crossing fan
A
projection of the area: Exercise: Use the geometrical interpretation to find the electric flux in the following situations:
E E n A A┴ =Acosθ
Flux: Ф=0 Ф=Ф0cosθ Ф=E A┴ 6 Electric Flux. Geometrical meaning. Geometrical
Closed surface.
point charge inside sphere put put one more charge Q move second Q off center off Q 2Q Q Q Flux Ф Flux 2Ф Flux 2Ф replace the sphere with an arbitrary closed surface Q
Flux Ф 7 Electric Electric Flux. Geometrical meaning. Geometrical
Closed surface.
place the charge outside the closed surface ? Flux = 0 every line that goes in  goes out Suppose the game got cancelled Every fan enters the stadium, Gets a rain check And leaves through another exit.
0 Total human flux = ?
8 Gauss’ Gauss’ Law.
From the previous slide, Ф through a closed surface:
• • • • Does NOT depend on the location of the charge inside the surface Does NOT depend on the shape of the surface Equals zero if there is no charge inside the surface In fact Gauss’ Law: electric flux through an arbitrary electric (Gaussian) surface is proportional to the net charge enclosed by the surface. 9 Gauss’ Gauss’ Law at work. Strategy. Strategy.
Gauss’ Law helps to find the electric field created at symmetric some point P by a symmetric charge distribution. 1. Choose a Gaussian surface S
• through point P • shape dictated by the symmetry 2. Simplify the lux integral 3. Compute the net charge enclosed by S and use: 10 Gauss’ Gauss’ Law at work. Sample problems. Sample
1. A thin hollow sphere of radius a has total charge Q. Find the electric field created by the sphere at an arbitrary distance from its center r. Assume that charge is spread uniformly on the sphere. Q r a P Electric field outside the sphere does not depend on the charge distribution! (as long as it is spherically symmetric) Same (outside) field for a point charge, solid sphere, spherical shell etc. Electric field inside the sphere is ZERO! (does not depend on what is outside) 11 Gauss’ Gauss’ Law at work. Faraday Cage. Faraday
Michael Faraday 17911867
* Inventor of field lines * Unit of capacitance * Electromagnetic induction Do not worry! He was not imprisoned. He used a conducting cage to block external electric field. See the demo. (more details on the next recitation)
12 Gauss’ Gauss’ Law at work. Sample problems. Sample
2. Find the electric field created by a very long charged rod at an arbitrary distance from its axis r. Assume that the rod has constant linear charge density λ. λ r P 3. AT HOME. Find the electric field created by a very large uniformly charged plane with surface charge density σ. 13 Gauss’ Gauss’ Law at work. Sample problems. Sample
4. Find the electric field created by a uniformly charged nonconducting solid sphere at an arbitrary distance from its axis r. Assume that the radius of the sphere is a, and the charge density is ρ. ρ a r P A spherical void of radius b was cut inside the sphere above. Find the electric field inside the void if its center is displaced by vector R. R
Hint: use the result of the previous problem and the superposition principle
14 Gauss’ Gauss’ Law at work. Summary. Summary.
Electric field outside a charge distribution
Type of symmetry (spherical) (cylindrical) (planar) Field behavior Electric field inside a charge distribution
Depends on the details of the distribution, e.g. • E=0 for an empty or metal sphere • E=4πkρr/3 for a uniformly charged solid sphere
15 Gauss’ Gauss’ Law at work. Challenge problems. Challenge
4. Consider a uniformly charged ring, as shown in Fig 1. Find the direction of the electric field at point P (or prove that it is zero). Hint: The Gaussian surface in Fig. 2, which depicts a vertical diametric crosssection of the ring passing through point P, could be useful. Top view P Fig. 1 5.
P Gaussian surface Side view ring Fig. 2 Charge Q is placed at a vertex of a cube with side length a. What is the electric flux through each face of the cube? Q 16 Next Next time. 1. Work done by electric field 2. 2. Electric potential energy 3. Electric Potential 17 ...
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 Spring '10
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 Charge, Gauss' Law

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