Correct Answer - a. `q_A = -Q/(3) ; q_B = Q/(3)`
b. `V_A = Q/(16 pi epsilon_0 R) ; V_B (5 Q)/(48 pi epsilon_0 R)`
a. Let the charges on `A` and `C` be `q_1` and `q_2`, respectively (Fig. S3. 21). From conservation of charge, we have `q_1 + q_2 = 0`.
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Hence, `q_1 = -q_2`. Since `A` and `C` are connected by a conducting wire, so they have the same potential.
`V_A` = potential of A
=`(1)/(4 pi epsilon_0) q_(1)/(R) + (1)/(4 pi epsilon_0) Q/(2 R) + (1)/(4 pi epsilon_0) q_(2)/(4 R)`
`V_C` = potential of C
=`(1)/(4 pi epsilon_0) q_(1)/(4 R) + (1)/(4 pi epsilon_0) Q/(4 R) + (1)/ (4 pi epsilon_0) q_(2)/(4 R)`
Equalizing the potentials of `A` and `C`, i.e., `V_A = V_C`, we get
`q_(1)/( R) + Q/(2 R) + q_(2)/(4 R) = q_(1)/(4 R) + Q/(4 R) + q_(2)/(4 R)`
or `4 q_1 + 2 Q = q_1 + Q`
or `q_1 = -Q//3`
Hence, `q_2 = Q//3`.
b. `V_A = (1)/(4 pi epsilon_0 R) [(-Q)/(3) + Q/(2) + Q/(12)] = Q/(16 pin epsilon_0 R)`
`V_B = (1)/(4 pi epsilon_0) q_(1)/(2 R) + (1)/(4 pi epsilon_0) Q/(2 R) + (1)/ (4 pi epsilon_0) q_(2)/(4 R)`
=`(1)/(8 pi epsilon_0 R)[(-Q)/(3) + Q + Q/(6)] = (5 Q)/(48 pi epsilon_0 R)`.