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A insulating cylindrical rod of diameter `d` and length `l(l gt gt d)` has a uniform surface charge density such that the electric field just outside the curved surface of the cylinder at point `M` is `E_(0)`. Find the electric field due to charge distribuition at point `P (r gt gt l)`. A. `E_(0)(ld)/(2r^(2))`B. `E_(0)(ld)/(4r^(2))`C. `E_(0)(ld)/(3r^(2))`D. `E_(0)(2ld)/(r^(2))` |
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Answer» `(2lambda)/(2pi epsilon_(0)d)=E_(0)` `lambda=pi epsilon_(0)dE_(0)` So, `E_(p)=(E_(0)dl)/(4r^(2))` |
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