There are four concentric shells A,B, C and D of radii `a,2a,3a` and `4a` respectively. Shells B and D are given charges `+q` and `-q` respectively. Shell C is now earthed. The potential difference `V_A-V_C` is `k=(1/(4piepsilon_0))`

Explanation

Let shell C acquires charge 'q' which will be such that final potential of C is zero.
`V_(c)=(kq)/(3a)+(kq')/(3a)+((-kq)/(4a))=0`
`(kq)/(3a)+(kq')/(3a)=(kq)/(4a)impliesq'=3q((1)/(4)-(1)/(3))`
`q'=-(q)/(4)`

`V_(c)=0`
`V_(A)-V_(C)=V_(A)`
Now calculating `V_(A)` we get
`V_(A)=(kq)/(2a)-(k(q//4))/(3a)-(kq)/(4a)`
`V_(A)=(kq)/(6a)`
`V_(A)-V_(C)=(kq)/(6a)`