First slide

Gauss's law and its applications

Question

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

Easy

A

zero
B

$q / ε_{0}$
C

$- q / ε_{0}$
D

none of these

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/ $ε$ ₀.

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