A sphere of 4 cm radius is suspended with in a hollow sphere of 6 cm radius. If the inner sphere is charged to a potential 3 esu while the outer sphere is earthed, then the charge on the inner sphere will be

Explanation

Fist of all understand that ESU means electrostatic unit ( cgs unit ) and in this system `(1)/(4piepsilon_(0))` is replaced by 1. So potential due to Q esu of charge will be Q/R in ESU . Now the diagram for the above question is as follows:

Let the inner sphere & outer sphere have charges `Q_(1)` and `Q_(2)` on them respectively. Their radii being `R_(1)` and `R_(2)` such that `(R_(1) lt R_(2))` . Given `R_(1) =4 cm` and `R_(2)=6 cm ` . We have two unknowns `Q_(1)` and Consider the outer sphere . It is grounded so the potential of this spheres is zero .
or `(Q_(1))/(R_(2))+(Q_(2))/(R_(2)) cdots (i)`
Note the contribution of `Q_(2)` in potential will be its surface potential but for `Q_(1)` the outer sphere is outside it so `Q_(1)` will be treated at the center .
Hence from equation (1) `Q_(1)+Q_(2) =0 implies Q_(2)=-Q_(1)`

Now write the expression for the potential of inner sphere and make it equal to 3 esu. For the inner sphere both the charges will contribute their respective surface potentials .
Therefore `(Q_(1))/(R_(1))+(Q_(2))/(R_(2))=3`
Substituting `Q_(2)=-Q_(1)` we get
`(Q_(1))/(R_(1))-(Q_(1))/(R_(2))=3` therefore `Q_(1) [(1)/(4)-(1)/(6)]=3 ` , or `Q_(1)=36` esu.