Charge on an originally uncharged conductor is separated by holding a positively charged rod very nearby, as…
Charge on an originally uncharged conductor is separated by holding a positively charged rod very nearby, as in figure Assume that the induced negative charge on the conductor is equal to the positive charge \(q\) on the rod. Then, flux through surface \(S_{1}\) is
zero
\(q / \varepsilon_{0}\)
\(-q / \varepsilon_{0}\)
none of these
Solution
Net charge on the conductor will be zero. So, net charge inside \(S_{1}\) will be the charge on the rod. hence, flux through \(S_{1}\) is \(q / \varepsilon_{0}\)
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Asked in: JEE Mains - Electrostatics - Chapter Test
