Charge on an originally uncharged conductor is separated by holding a positively charged rod very closely nearby, as shown in figure. Assume that the induced negative charge on the conductor is equal to the positive charge q on the rod. Then the flux through surface S1 is
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a
zero
b
q/ε0
c
−q/ε0
d
none of these
answer is B.
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Detailed Solution
Net charge on the conductor will be zero. So, net charge inside ,S1 will be the charge on the rod. Hence, flux through S1 is q/ε0.