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=4sigma pi r_(0)^(2)`
B. `r_(0)=lambda/(2pi sigma)`
C. `E_(1)(r_(0)//2)=2E_(2)(r_(0)//2)`
D. `E_(2)(r_(0)//2)=4E_(3)(r_(0)//2)`

Answer

C

Explanation

Point charge `E_(1)(r_(0))=Q/(4pi in_(0)r_(0)^(2))`
Line charge `E_(2)(r_(0))=lambda/(2pi in_(0)r_(0))`
Infinite sheet `E_(3)(r_(0))=sigma/(2in_(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/(2 in_(0))` ...(i)
So, `Q/(4pi in_(0)r_(0)^(2))=lambda/(2pi in_(0)r_(0))rArr Q=2lambdar_(0)`
Now, `E_(1)(r_(0)//2)=Q/(4pi in_(0)(r_(0)/2)^(2))=Q/(pi in_(0)r_(0)^(2))`
`=(2lambdar_(0))/(pi in_(0) r_(0)^(2))=(2lambda)/(pi in_(0) r_(0))`
`E_(2)(r_(0)//2)=lambda/(2pi in_(0) r_(0)/2)=lambda/(pi in_(0) r_(0))`
`=(E_(1)(r_(0)//2))/2 ne 4 E_(3)`
From equation (i)
`Q/(4pi in_(0)r_(0)^(2))=sigma/(2 in_(0))rArr Q=2sigma pi r_(0)^(2)`
Also from equation (i)
`lambda/(2pi in_(0)r_(0))=sigma/(2 in_(0))rArr r_(0)=lambda/(sigma pi)`