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Two concentric shells of radii R and 2R ...

Two concentric shells of radii R and 2R are shown in the figure. Initially a charge q is imparted to the inner shell. After the keys `K_(1)&K_(2)` are alternately closed n times each, find the potential difference between the shells.

Text Solution

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When `K_(1)` is closed first time, outer sphere is earthed and its potential becomes zero. Let the charge on it be `q_(1)`.
`V_(1)`'= Potential due to charge on inner sphere and that due to charge on outer sphere.
`0=(1)/(4piepsilon_(0))[(q)/(2R)+(q_(1))/(2R)]`
`implies q_(1)'=q`
When `K_(2)` is closed first time, the potential `V_(2)'` on inner sphere becomes zero as it is earthed.
Let the new charge on inner sphere be `q_(2)`'
`0=(1)/(4piepsilon_(0))(q_(2))/(R)+(1)/(4piepsilon_(0))((-q))/((2R))`
`implies q_(2)'=q//2`
Now when `K_(1)` will be closed second time charge on outer sphere will be `-q_(2)`' i.e. `-q//2`
Similarly when `K_(1)` will be closed nth time, charge on outer sphere will be `-(q)/(2^(n-1))` as each time.
Charge will be reduced to half the previous value.
After closing `K_(2)` nth time charge on inner shell will be negative to half the charge on outer shell.
i.e. `(-(q)/(2^(n)))` and potential on it will be zero.
For potential of outer shell
`V_(0)=(1)/(4piepsilon_(0))((-q//2^(n)))/(2R)+(1)/(4piepsilon_(0))(((q)/(2^(n-1))))/(2R)`
`V_(0)=(q[-1+2])/(4piepsilon_(0)2^(n+1)R)=+(q)/(4piepsilon_(0)2^(n+1)R)`
Potential difference = `V_(0)-V_(1)=(q)/(4piepsilon_(0)2^(n+1)R)-0=(q)/(4piepsilon_(0)2^(n+1)R)`
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