InterviewSolution
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InterviewSolution
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Suposse the surface charge density over a sphere of radius R dendendson a polarangletheta as sigma = sigma_(0) cos theta, where sigma_(0) is a positiveconstant. Show that such a charge distribution can be represented as a resultof a small relative shift of two uniformly charge balls of radiusR whosechargesare equalin magnitudeand oppositein sign. Restoringto thisrepresentation, find teh electricfield strength vector inside the given spehre. |
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Answer» Solution :We START from two charged SPHERICAL balls each of radius `R` with equaland opposite charge densities`+ rho` and `-rho`. The CENTRE of the balls are at `+ (vec(a))/(2)` and `- (vec(a))/(2)` respectively so the equaction of their surfaces are `|vec(r ) - (vec(a))/(2)| = R` or`r - (a)/(2) cos theta= R` and `r + (a)/(2) cos theta = R`, considering `a` to be small. The distance between the two surface in the radial direction at angle `theta` is`|acos theta|` and does not dependon TEH azimuhal angle. It is seen from the diagram that the surface of the sphere has in effect a surface density `sigma = sigma_(0) cos theta` when `sigma_(0) = rho a`. Insideany uniformlychargedspherical ball, the field is radial and has the magnitudegiven by Gaussain's theorem `4pi r^(2) E = (4pi)/(3) r^(3) rho//epsilon_(0)` or`E = (rho r)/(3 epsilon_(0)` In vector notation, using the fact the `V` myst be measured from the centre of the ball, we get, for the present case `vec(E) = (rho)/(3 epsilon_(0)) (vec(r ) - (a)/(2)) - (rho)/(3 epsilon_(0)) (vec(r ) + (vec(a))/(2))` `= -rho vec(a)//3 epsilon_(0) = (sigma_(0))/(3 epsilon_(0)) vec(k)` When `vec(k)` is teh unit vector along the polar axis from which`theta` is measured. 
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