Electric flux To state Gauss’s Law in a quantitative form, we first need to define Electric Flux. # of field lines N = density of field lines x “area” where “area” = A2 x cos  

Why are we interested in electric flux? is closely related to the charge(s) which cause it. Consider Point charge Q If we now turn to our previous discussion and use the analogy to the number of field lines, then the flux should be the same even when the surface is deformed. Thus should only depend on Q enclosed.

Gauss’s Law: Quantitative Statement The net electric flux through any closed surface equals the net charge enclosed by that surface divided by 0. How do we use this equation?? The above equation is TRUE always but it doesn’t look easy to use. BUT - It is very useful in finding E when the physical situation exhibits a lot of SYMMETRY.

READING QUIZ 1 Consider a spherical conducting shell with an outer radius b and inner radius a. The shell is insulated from its surroundings and has a net charge of +Q. A point charge –q is inserted into the shell without touching the shell and placed at r = a/2. Which statement is correct ? A| The net charge on the outer surface does not change. B| A charge –q is induced on the inner surface of the shell. C| When the point charge is moved from r = a/2 to r = 0 the charge distribution on the outer surface does not change. D| The net charge on the outside of the shell is Q+q

Infinitely long uniformly charged line Gauss’s Law: Same result but much less work!

Uniformly charged thin, infinite sheet Gauss’s Law! A h If E=0 on one side, then what is E between the plates?

Thin sheet of any charge distribution tiny disk Just to right of disk Just to left of disk

Physics 241 –Warm-up quiz The magnitude of the electric field from a single charged plane of charge density  and infinite area is E = 0). What is the magnitude of the field at point A,B and C ? Note that the planes are infinitely large.  - (A). E=0 in A and C but E= to the left in B (B). E=2 to the left in A and C but E=0 in B. (C). E=2 to the left in A and C but E=  in B. (D). E=0 in A and C but E= to the right in B A B C

Charges and fields of a conductor In electrostatic equilibrium, free charges inside a conductor do not move. Thus, E = 0 everywhere in the interior of a conductor. Since E = 0 inside, there are no net charges anywhere in the interior. Net charges can only be on the surface(s). The electric field must be perpendicular to the surface just outside a conductor, since, otherwise, there would be currents flowing along the surface.

DOCCAM 2

Electrostatic Shielding (Faraday Cage) E=0 in empty cavity in a good conductor. Why? If there are E field lines, then moving a charge from one end to the other against the E line through the cavity must require some work W. On the other hand since E is perpendicular to surface, W=0 if you move the charge along the surface instead. But the work should be independent of the path used. => Contradiction! X

DOCCAM 2

Electric Field of a Conductor: Examples Net charge Q ( > 0) on the conductor Hollow region Net charge induced = +q Hollow region Field ? Net charge –q < 0 + Net charge (Q-q) on the outside surface

Cavity in a spherical conductor With +q in the cavity of a conductor, a net charge –q is induced on the inside surface of the cavity. If +q is at the center of the spherical cavity, E is radial by symmetry. Uniform charge distributions. Total induced charge -q As long as the E contribution due to all the interior charges, measured at the outer surface is zero, then moving the +q charge does not affect the exterior surface charge distribution. Electrostatic shielding

Electrostatic Shielding (Continued) If you move charge q in the cavity, the exterior electric fields and the exterior charge distribution are not affected. q Conducting shell electrostatically shields its exterior from changes on the inside. If you now add charge Q’ to the conductor and/or Q’’ on the outside of the conductor, the interior electric fields do not change. Q’’ Add Q’ + Conducting shell electrostatically shields its interior from changes on the outside, too.

DOCCAM 2

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Two conducting plates In isolation, the electric field near a charged conducting plate is given by properties of conductors and, superposition What happens if these two plates are pulled toward each other?

Two conducting plates continued The excess charge on one plate attracts the excess charge on the other plate. The surface charge density vanishes on the outer faces but is doubled on the inner faces. The magnitude of the electric field is twice as much as that near an isolated plate and is zero everywhere else. Case (c) is not a simple superposition of cases (a) and (b), because the electric charges are re-distributed on both plates when they are brought near each other.

Physics 241 – 10:30 Quiz 2 The magnitude of the electric field from a single charged plane of charge density  and infinite area is E = 0). What is the magnitude of the field at point A on the right side of the two infinitely large planes of uniformly charged?  - to the right 3to the left to the right (2)to the left 3to the right A e = 1.610-19 C  = 8.8510-12 C2/(Nm2)

Physics 241 – 11:30 Quiz 2 The magnitude of the electric field from a single charged plane of charge density  and infinite area is E = 0). What is the magnitude of the field at point A between the two infinitely large planes of uniformly charged?   to the right 3to the left to the right (2) to the left to the left A e = 1.610-19 C  = 8.8510-12 C2/(Nm2)

Physics 241 –Quiz 3c The magnitude of the electric field from a single charged plane of charge density  and infinite area is E = 0). What is the magnitude of the field at point A on the left side of the two infinitely large planes of uniformly charged?   to the right 3to the left to the right (2)to the left 3to the right A e = 1.610-19 C  = 8.8510-12 C2/(Nm2)