1.

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).

Answer»

<P>`E_(0)(ld)/(2R^(2))`
`E_(0)(ld)/(4r^(2))`
`E_(0)(ld)/(3r^(2))`
`E_(0)(2ld)/(r^(2))`

Solution :`(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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