Explanation:
It is given that,
Charge, [tex]Q_1=5.7\ nC=5.7\times 10^{-9}\ C[/tex]
Charge, [tex]Q_2=-2.3\ nC=-2.3\times 10^{-9}\ C[/tex]
Distance between charges, d = 45 cm = 0.45 m
We need to find the electric potential at a point midway between the charges. It is given by :
[tex]V=\dfrac{kQ_1}{r_1}+\dfrac{kQ_2}{r_2}[/tex]
[tex]V=k(\dfrac{Q_1}{r_1}+\dfrac{Q_2}{r_2})[/tex]
Midway between the charges means, r = 22.5 cm = 0.225 m
[tex]V=9\times 10^9\times (\dfrac{5.7\times 10^{-9}\ C}{0.225\ m}+\dfrac{-2.3\times 10^{-9}\ C}{0.225\ m})[/tex]
[tex]V=136\ volts[/tex]
So, the electric potential at a point midway between the charges is 136 volts. Hence, this is the required solution.
