The potential at a point, due to a positive charge of $100\,\mu C$ at a distance of $9\,m$, is
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(b) By using $V = 9 \times {10^9} \times \frac{Q}{r}$$ = 9 \times {10^9} \times \frac{{100 \times {{10}^{ - 6}}}}{9} = {10^5}\,V$
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