Phys. 122: Thursday, 01 Oct. HW 4 returned: please pick up yours in front. Mast. Phys.: Assign. 4 due this evening. Assign. 5 due in one week. Written HW 6: due Tuesday. Reading: Should be done with ch. 28. Read ch. 29 by next Thursday. Exam 1: is still being graded; we'll likely go over it on Tuesday.

If E dot n (or E dot dA in your book) is known by symmetry to be constant on the surface, it factors out of the integral. (This is the most useful case of Gauss' Law, as it lets you find E from where the charges are.) Then, (E dot n)(Area) is the value of the flux on the left-hand side above. Any closed surface which lies entirely on the interior of a conductor must have zero total charge inside! (second most useful)

Clicker Question 2 A spherical surface surrounds an isolated positive charge. We can calculate the electric flux for this surface. If a second charge is placed outside the spherical surface, what happens to the magnitude of the flux? A.The flux decreases proportionally to the magnitude of the second charge. A.The flux does not change. A.The answer depends on whether the second charge is positive or negative. A.The flux increases proportionally to the magnitude of the second charge.

Clickers: A negative point charge is inside a uniform spherical shell of positive charge, but it's off to one side and not centered. The net force on it will be... a) Away from the near side (toward the center of the shell) b) Toward the near side (away from the center) c) Directly upward d) Directly downward e) Zero

Spherically Symmetric Cases A case is spherically symmetric if the E field can only point in the (plus or minus) r direction, and can only depend upon the distance r (and not the direction) In that case, the flux is always |E| times 4 π r ² (the surface area of a sphere). The E field looks like the field of a point charge, but with the total interior charge Q as the source. Useful for: point charges, spherical conductors, spherical charged insulators, etc.

R 1 R 2 R 3 Charge Q 1 on inner conductor; Q 2 total on outer

Clickers: A uniform flat sheet of charge with charge per area σ creates an electric field around it. If the same charge density is on a conductor's surface, how strong is its E field just outside, in comparison with the charged sheet? a) Twice as strong b) Half as strong c) The same strength d) Depends upon the conductor's shape e) E must be zero outside a conductor

For ANY CONDUCTOR in equilibrium: very close to the surface, |E| (just outside) is related to σ.

The Electric Potential (or just Voltage) gives the potential energy per charge:...turned around, this becomes (Δ U) = q ( Δ V).

Clickers: the units of electric potential are... a) Newtons per Coulomb b) Coulombs per Newton c) Volts per meter d) Joules per Coulomb e) Joules per second

Uses for the potential difference: Conservation of energy: differences in V for a charged particle cause differences in the potential energy U, which makes the kinetic energy change. Finding the electric field E: it's usually much easier to find V first, and use that to find E.

Clickers: A particle with positive charge travelling with speed v goes from a region of large electric potential V to a region of lower V. What will happen to the charge's speed v? (a) It will increase. (b) It will decrease. (c) It will stay the same. (d) It will reverse. (e) Not enough information is given to decide.

The Electric Potential (or just Voltage) gives the potential energy per charge:...turned around, this becomes (Δ U) = q ( Δ V).

This fact below shows that electrostatic forces are conservative. Recall from Phys. 121 that only conservative forces have potential energies!