Four concentric hollow spheres of radii R, 2R, 3R, and 4R are given the charges as shown in figure. Then the conductors 1 and 3, 2 and 4 are connected by conducting wires (both the connections are made at the same time.). The final charge on conductor 4 is-

Options

A. `22Q//5`
B. `+11Q//3`
C. `-11Q//3`
D. `-22Q//5`

Answer

D

Explanation

Let the charge distribution be as shown in fig
From gauss's theorem we know that facing surface s of the conductor acquire equal and oppsite charges. So
`V_(1)=V_(3)` and `V_(2)=V_(4)`
`q_(1)+q_(3)-q_(2)=+4Q` .(i)
Now, `q_(2)-q_(1)+q_(4)-q_(3)=-6Q` ..(ii)
`V_(1)=(1)/(4piepsilon_(0))[(q_(1))/(R)+(q_(2)-q_(1))/(2R)+(q_(3)-q_(2))/(3R)+(q_(4)-q_(3))/(4R)]`
`V_(2)=(1)/(4piepsilon_(0)[(q_(1))/(2R)+(q_(2)-q_(1))/(2R)+(q_(3)-q_(2))/(3R)+(q_(4)-q_(3))/(4R)]`
`V_(3)=(1)/(4piepsilon_(0))[(q_(1))/(3R)+(q_(2)-q_(1))/(3R)+(q_(3)-q_(2))/(3R)+(q_(4)-q_(3))/(4R)]`
`V_(4)=(1)/(4piepsilon_(0))[(q_(1))/(4R)+(q_(2)-q_(1))/(4R)+(q_(3)-q_(2))/(3R)+(q_(4)-q_(3))/(4R)]`
from `V_(1)=V_(3),q_(1)=-(q_(2))/(3)`
from `V_(2)=V_(4),q_(2)=-(q_(3))/(2)`
On solving eq. (ii) `q_(1)=(2Q)/(5),q_(2)=-(6Q)/(5)` and `q_(3)=(12Q)/(5)`
Substituting these values in Eq (ii) we get `q_(4)=-2Q`
charge on the inner surface of the third conductor is `-q_(2)=6Q//5`
charge on the fouth conductor is `q_(4)-q_(3)=-2Q-(12Q)/(5)=(-22Q)/(5)`
potential of conduction 1, `V_(1)=(-3Q)/(40piepsilon_(0)R)=V_(3)`
Potential of conductor 2, `V_(2)=(Q)/(8piepsilon_(0)R)=V_(4)`