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A point charge q is locatedin vacumm at a distancel fromthe planesurfaceof a unifrom isotropicdeilectric filling up all the half-space. The permittivity of the dielectricequals epsilon. Find : (a) the surfacedensityof the bound chagresas a function of distanc e r from the pointcharges q'. anayse the obtainedresultat l -. 0, (b) the total bound charge on the surfaceof the dielectric |
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Answer» Solution :The charge is at `A` in the medium 1 and has an image point at `A'` in the medium 2. The electric FIELD in themedium 1 is due to the actual charge `q` at `A` and the image chagre `q'` at `A'`. The electric field in 2 is due to a connectedcharge `q'` at `A`. Thus on the boundary between 1 and 2. `E_(1n)= (q')/(4pi epsilon_(0) R^(2)) cos theta - (q)/(4pi epsilon_(0) r^(2)) cos theta` `E_(2n)= (-q'')/(4pi epsilon_(0) r^(2)) cos theta` `E_(1t)= (q')/(4pi epsilon_(0) r^(2)) sin theta + (q)/(4pi epsilon_(0) r^(2)) sin theta` `E_(2T)= (q'')/(4pi epsilon_(0) r^(2)) sin theta` The boundary conditions are `D_(1n) = D_(2n)` and `E_(1t) - E_(2t)` `eq" = q - q` `q" = q + q` So, `q" = (2q)/(epsilon + 1), q' = (epsilon - 1)/(epsilon + 1) q` (a) The surface DENSITYOF the round charge on the surface of the dielectric `sigma' = P_(2n) =D_(2n) - epsilon_(0) E_(2n) = (epsilon - 1) E_(0) E_(2n)` `=(epsilon -1)/(epsilon +1)(q)/(2pi r^(2)) cos theta = (epsilon - 1)/(epsilon +1) (ql)/(2pi r^(3))` (B) Total bound chargeis, `(epsilon - 1)/(epsilon +1) q int_(0)^(oo) (l)/(2pi (l^(2) + x^(2))^(3//2)) 2pi x dx = - (epsilon - 1)/(epsilon + 1) q` 
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