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There are two spheres of radii R and 2R ...

There are two spheres of radii `R` and `2R` having charges `Q` and `Q//2`, respectively. These two spheres are connected with a cell of emf `V` volts as show in. When the switch is closed, find the final charge on each sphere.

Text Solution

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When the switch is closed, the potential difference between the spheres should be `V`. Let `q` charges flow from the sphere of radius `R`.
`((q_(1))_("final"))/C_(1)=((q_(2))_("final"))/C_(2)=V`
`C_(1)=4piepsilon_(0)(2R), C_(2)=4piepsilon_(0)R`
Then
`((Q/2+q))/(4piepsilon_(0)(2R))-((Q-q))/(4piepsilon_(0)(R))=V`
or `Q/2+q-2Q+2Q=4piepsilon_(0)(2R)V`
or `q=(8piepsilon_(0)RV)/3+Q/2`
So the final charges on each of the spheres are
`Q_(1)=Q-q=(-8piepsilon_(0)RV)/3+Q/2`
and `Q_(2)=Q/2+q=Q+(8piepsilon_(0)RV)/3`.
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