Let `E_1(r)`, `E_2(r)` and `E_3(r)` be the respectively electric field at a distance r from a point charge `Q`, an infinitely long wire with constant linear charge density `lambda`, and an infinite plane with uniform surface charge density `sigma`. If `E_1(r_0)=E_2(r_0)=E_3(r_0)` at a given distance `r_0`, then

Options

A. `Q = 4 sigma pi r_(0)^(2)`
B. `r_(0) = lambda//2 pi sigma`
C. `E_(1) (r_(0)//2) = 2 E_(2)//(r_(0)//2)`
D. `E_(2) (r_(0)//2) = 4 E_(0) (r_(0)//2)`

Answer

C

Explanation

Here, `E_(1) (r_(0)) = (Q)/(4pi in_(0) r_(0)^(2)) , E_(1) (r_(0)//2) = (4Q)/(4pi in_(0) r_(0)^(2))`
`E_(2) (r_(0)) = (lambda)/(2pi in_(0) r_(0)) , E_(2) (r_(0)//2) = (lambda)/(pi in_(0) r_(0))`
`E_(3) (r_(0)) = (sigma)/(2in_(0)) , E_(3) (r_(0)//2) = (sigma)/(2 in_(0))`
Given, `E_(1) (r_(0)) = E_(2) (r_(0)) = E_(3) (r_(0))`
`:. (Q)/(4pi in_(0) r_(0)^(2)) = (lambda)/(2pi in_(0) r_(0)) = (sigma)/(2in_(0))`
Hence, `Q = 2pi r_(0)^(2) sigma`
and `(lambda)/(pi r_(0)) = sigma` or `r_(0) = (lambda)/(pi sigma)`
`E_(1) ((r_(0))/(2)) = 4E_(1) (r_(0)) = 2 E_(2) (r_(0)//8)`
Also `E_(2) ((r_(0))/(2)) = 2 E_(0) (r_(0)) = 2 E_(3) (r_(0)//2)`