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A cylindrical portion of radius r is removed from a solid sphere of radius R and uniform volume chargedensity `rho` in such a way that the axis of the hollow cylinder coincides with one of the diameters of thesphere. (r is negligible compared to R). Then the electric field intensity at point A isA. `(rhor)/(3epsilon_(0))hati`B. `-(rhor)/(3epsilon_(0))hati`C. `(rhor)/(6epsilon_(0))hati`D. `-(rhor)/(6epsilon_(0))hati` |
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Answer» Field at A Due at A due the cylinder of length 2R (which can assumed to be inifinite, since `r lt lt R`) `E_(2)=(2K(rhopir^(2)))/(r)(-hati)=-(rho)/(2epsilon_(0))rhati` `therefore` net field `E=E_(1)+E_(2)=(rhor)/(6epsilon_(0))hati` |
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