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A uniformly charged and infinitely long ...

A uniformly charged and infinitely long line having a linear charge density `'lambda'` is placed at a normal distance `y` from a point `O`. Consider a sphere of radius `R` with `O` as centre and `Rgty`. Electric flux through the surface of the sphere is

A

zero

B

`(2lambdaR)/(epsilon_(0))`

C

`(2lambdasqrt(R^(2)-y^(2)))/(epsilon_(0))`

D

`(lambdasqrt(R^(2)+y^(2)))/(epsilon_(0))`

Text Solution

Verified by Experts

The correct Answer is:
C
Electric flux `oint_(s) vecE.vec(dS)=(q_("in"))/(epsilon_(0)) " "q_("in")` is the charge enclosed by the Gaussian-surface which, in the present case, is the surface of given sphere. As shown, length AB of the line lies inside the sphere.

In `Delta OO'A" " R^(2) =y^(2) +(O'A)^(2)`
`therefore O'A=sqrt(R^(2)-y^(2))`
and `AB=2sqrt(R^(2)-y^(2))`
Charge on length `AB=2 sqrt(R^(2)-y^(2)) xx lambda`
`:.` electric flux `= oint_(s) vecE. vec(dS)= (2lambda sqrt(R^(2)-y^(2)))/(epsilon_(0))`
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