4. Gauss’s law

1. 4. Gauss’s law 4.1 Electric flux • Uniform electric field - number of electric field lines Perpendicular to the area area • Not uniform electric field Definition: Units:

2. Example: Flux through open surface Compare the electric flux through two squares of areas A1 and A2 (A2>A1) placed in a region with a uniform electric field E: A2 θ θ A2 cosθ (Side view) • Φ1 < Φ2 • Φ1 = Φ2 • C. Φ1 > Φ2 E A1 A2 The number of lines going through them is the same. The effective area for the large square is A2cosθ = A1 • If you reverse the direction of A, you reverse the sign of the flux • For an open surface, choose any direction • For a closed surface it is conventional to take the area vector pointing outwards • Ф> 0  lines going out • Ф < 0  lines coming in • Ф = 0  no lines or a balance between incoming and outgoing lines

3. Example 1: Flux through closed surface A cubic box is placed in a region of uniform electric field as shown in figure 1. If the cube is tilted, as shown in figure 2, the net electric flux through it: + 2R E R E • Increases • B. Decreases • C. Stays the same The net flux is zero in both cases! Every line that comes in, goes out. Example 2:Two spheres have radii R and 2R. Their centers lie on a positive charge. Compare the electric flux through the two surfaces: A.ΦR < Φ2R B. ΦR = Φ2R C. ΦR > Φ2R The number of lines going through both surfaces is the same. The area at 2R is 4 times larger, but the electric field is 4 times smaller!

4. 4.2 Gauss’s law - generalization of Coulomb’s law Gauss’s law: Example:

5. 4.3 Applications of Gauss’s law • Shell theorem: q r Q R • Cylindrical symmetry

6. Examples (applications of Gauss’s law): Electric field created by

7. Planar symmetry a) Conducting surface b) Conducted plate c) Two conducting plates Q +Q -Q Q/2+Q/2=Q +++++ +++++ ----- +++++ +++++ E E=0 E E=0 E=0 E E

8. Example: A solid metal sphere of radius 3.00 m carries a total charge of 3.5μC. What is the magnitude of the electric field at a distance from the sphere’s center of (a) 0.15 m, (b) 2.90 m, (c) 3.10 m, and (d) 6.00 m? (e) How would the answers differ if the sphere were a thin shell? • and (b) Inside a solid metal sphere the electric field is 0 (c) Outside a solid metal sphere the electric field is the same as if all the charge were concentrated at the center as a point charge: (d) Same reasoning as in part (c): (e) The answers would be no different for a thin metal shell

9. Example: Find the magnitude of the electric field produced by a uniformly charged sphere with radius R and total charge Q. E r R r R For r > R: For r < R:

10. Example: Find the magnitude of the electric field produced by a uniformly charged cylinder with radius R length L and total charge Q. E r R r R For r > R: For r < R:

11. A charged isolated conductor • The static electric field inside a conductor is zero – if it were not, the charges would move. • The net charge on a conductor in equilibrium is on its surface, because the electric field inside the conductor is zero. Spherical conducting shell: Gauss’s surface Charge inside the surface inside shell(+Q) inside conductor(+Q) +(-Q)=0 otside shell(+Q) +(-Q) + (+Q) = +Q

12. Example: Compare the magnitude of the electric field at point P before and after the shell is removed. P σout σin σin Q Q Hint: Use Gauss’s law to find EP. A. Ebefore < Eafter B. Ebefore = Eafter C. Ebefore > Eafter In both cases, the symmetry is the same. We will use the same Gaussian surface: a sphere that contains point P. So the flux will look just the same: And the enclosed charge is also the same: Q. Therefore,