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Unformatted text preview: HW-7 1.A sphere of radius R carries a polarization rkrP)(, where k is a constant. (a)Calculate the bound charge band b. (b)Find the field inside and outside the sphere. Solution:(a)At the surface of the sphere , kRRRknPb. Within the sphere, kzzkyykxxkzPyPxPPzyxb3. (b)From the symmetry, it is obvious that rrErE)()(. Then apply the Gauss’ law to a sphere of radius r, encQrrESdE24)(24)(rQrEencInside the sphere, 3334343krrkrdQbenc2344)(krrkrrEOutside the sphere, 4444423233RkRkRRkRQkRQbsurencWhich is expected because the total charge has to be zero. Then we haveE2.The space between the plate a parallel-plate capacitor is filled with two slabs of linear dielectric material. Each slab has thickness a, so the total distance between the plates is 2a. Slab 1 has a dielectric constant r1, and slab 2 has a dielectric constant r2. The free charge density on the top plate is and on the bottom plate is -. (a)Find the electric displacement in each slab. (b)Find the electric field in each slab. (c)Find the polarization in each slab. (d)Find the potential difference between the plates. (e)Find the location and amount of all bound charge. (f)Now that you know all the charge (free and bound), recalculate the field in each slab, and confirm your answer to (b). Solution:(a)From the result of capacitor without the dielectric material, it is easy to obtain DWith the direction from +to -....
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This note was uploaded on 02/06/2010 for the course PHYSICS 11 taught by Professor Qiu during the Fall '09 term at Berkeley.
- Fall '09