Electrostatic potential

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Question

The potential at a point due to a positive charge of 100 C at a distance of 9m is $10^{x}$ V. What is $x$
value.

Easy

Solution

$V = k \frac{Q}{r} = 9 \times 10^{9} \times \frac{100}{9} = 10^{11} V$

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