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  4. The potential at a point due to a positive charge of 100 μ C at a distance of 9 m, is

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Q. The potential at a point due to a positive charge of $100\, \mu C$ at a distance of $9\, m$, is

Electrostatic Potential and Capacitance

Solution:

$V=9 \times 10^{9} \times \frac{Q}{r}$
$ =9 \times 10^{9} \times \frac{100 \times 10^{-6}}{9}$
$=10^{5} V$