Correct Answer - A::B::C
Charge on Shell `A=q_A=sigma(4pia^2)`
Charge on Shell `B=q_B=sigma(4pib^2)`
Charge on Shell `C=q_C=sigma(4pic^2)`
The potential of shell A
`V_A=(kq_A)/(a)+(kq_B)/(b)+(kq_C)/(c)`
`=(ksigma(4pia^2))/(a)+(k(-sigma)(4pib^2))/(b)+(ksigma(4pic^2))/(c)`
`=(1)/(4piepsilon_0)xxsigmaxx(4pia^2)/(a)-(1)/(4piepsilon_0)sigma((4pib^2))/(b)+(1)/(4piepsilon_0)xxsigma((4pic^2))/(c)`
`=(sigma)/(epsilon_0)[a-b+c]` Similarly, `V_B=(kq_A)/(b)+(kq_B)/(b)+(kq_C)/(c)`
and `V_C=(kq_A)/(c)+(kq_B)/(c)+(kq_C)/(c)`
`V_B=(sigma)/(epsilon_0)[(a^2)/(b)-b+c]` and `V_C=(sigma)/(epsilon_0)[(a^2-b^2+c^2)/(c)]`
Given that `V_A=V_C`
`(sigma)/(epsilon_0)(a-b+c)=(sigma)/(epsilon_0)[(a^2-b^2+c^2)/(c)]`
or `ac-bc+c^2=a^2-b^2+c^2` or `c=a+b`
