Question

Two charges 2μC and 4μC are separated by a distance of 20cm in air. Find the electric potential at the midpoint of the line joining the two charges?​

Answer 1

Given ,

Two charges 2μC and 4μC are separated by a distance of 20 cm

We know that , The electric potential due to multiple charges is given by

$\large \sf \fbox{V = k( \frac{q_{1}}{r_{1}} + \frac{ q_{2} }{ r_{2}} + ... +\frac{ q_{n} }{ r_{n}} ) \: }$

Thus ,

$\sf \mapsto V = 9 \times {(10)}^{9} ( \frac{2 \times {(10)}^{-6} }{0.1} + \frac{4 \times {(10)}^{-6} }{0.1} ) \\ \\ \sf \mapsto V = \frac{9 \times {(10)}^{4} \times (2 + 4)}{1} \\ \\ \sf \mapsto V = 54 \times {(10)}^{4} \: \: volt$

$\therefore \sf \underline{The \: electric \: potential \: at \: the \: midpoint \: is \: 54 × {(10)}^{4} \: volt }$

Answer 2

Explanation:

In this way we can solve your problem my friend....