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Three conducting concentric shell s`A,B, and C` of radius `a,2a,3a` are as shown in the figure.The charge on shell `B` is `Q`..The switch `S` is closed.Then after closing the switch
If `sigma_(A),sigma_(B)` and `sigma_(C)` are the surface charge densities on `A,B and C` respectively then `sigma_(A):sigma(B):sigma(C)`is

A

`9:9:1`

B

`9:-9:1`

C

`-9:9:1`

D

`1:1:1`

Text Solution

Verified by Experts

The correct Answer is:
C
`sigma_(A)=(-Q)/(4(4pia^(2)))sigma_(B)=(Q)/(4pi4a^(2))sigma_(C)=(Q)/(44pi9a^(2)) sigma_(A): sigma_(B): sigma_(C)=-(1)/(4):(1)/(4):(1)/(36)=-9:9:1`
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