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Two charge d spherical conditioners of r...

Two charge d spherical conditioners of radii `R_(1)&R_(2)` when connented by a conducting wire acquire charge `q_(1)&q_(2)` respectively. Find the ratio of surface charge densities in terms of their radii.

Text Solution

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The charge will flow between the two spherical conditioners till their potentail become equal.
i.e. `(Kq_(1))/(R_(1))=(Kq_(2))/(R_(2))(or)(q_(1))/(q_(2))=(R_(1))/(R_(2))`
The ratio of the surface charge densities on the two conditioners will be
`(sigma_(1))/(sigma_(2))=(q_(1)//4piR_(1)^(2))/(q_(2)//4piR_(2)^(2))=(q_(1))/(q_(2)).(R_(2)^(2))/(R_(1)^(2))=(R_(1))/(R_(2))xx(R_(2)^(2))/(R_(1)^(2))`
`(sigma_(1))/(sigma_(2))=(R_(1))/(R_(2))`
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