Question

Charge on an originally uncharged conductor is separated by holding a positively charged rod very closely nearly, as in Fig. 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

• A. Zero
• B. $q/{\epsilon }_0$
• C. $-q/{\epsilon }_0$
• D. None of these

Answer 1

Answer: (B)

Initial charge on conductor $=0$
Due to charge rod, negative and positive charge will induce at near and far point of the rod.

$\therefore$ Induced positive charge $=$-(induced negative charge)
$=q$
On surface $S_1$, applying gauss' law :
${\int }_{s_1}\overrightarrow{E}.\overrightarrow{ds}=\frac{Qenc}{E_o}$
$\therefore$ Flux through $S_1=\frac{Totalchargedenclosed}{E_o}$
$=\frac{q+(q-q)}{E_o}$
$=\frac{q}{E_o}$