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Unformatted text preview: Fall 2009 - Lee - Midterm 2 solutions Problem 1 Solutions Part A Because the middle slab is a conductor, the electric field inside of the slab must be 0. Parts B and C Recall that to find the electric field near a plane charge of charge density σ , you can draw a Gaussian pillbox through it. Because the electric field points perpendicular to the plane, we only have a flux through the ends of the box. (See example 22-7 in the book, page 598) I-→ E ·-→ dA = 2 EA = σA E = σ 2 The first equality in the first equation is because the electric field is perpendicular and uniform to the ends of the box, and the second is Gauss’s law. (Note that A here is the area of the box, not of the plates) Because we have symmetry, all charge distributions will be sheet charges. If we define the positive direction to the right, and watch our sign conventions, E l (the electric field between the left plate and the middle slab) and E r (the electric field between the middle slab and the right plate) are: E l = σ 2- σ l 2- σ r 2-- σ 2 E r = σ 2 + σ l 2 + σ r 2-- σ 2 Where σ l and σ r are the charge densities on the left and right side of the conducting slab. Since the slab is neutral, these two charge densities are equal and opposite, so the middle two terms will add to 0 in both cases. Thus: E l = E r = σ Parts D and E Now we draw a pillbox with one end in the space between the slab and plate, and one end in the middle of the slab. Since the electric field inside the conductor is 0: I-→ E ·-→ dA =- EA = σ l I-→ E ·-→ dA = EA = σ r Note that the flux is negative if the box contains the left surface of the slab, and is positive if the box contains the right surface of the slab. Then, using our result from parts B and C for electric field, we find σ l =- σ σ r = σ 1 Problem 2...
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This note was uploaded on 03/17/2010 for the course PHYSICS ?? taught by Professor Zettle during the Fall '08 term at University of California, Berkeley.
- Fall '08