Answer:
Electric potential, [tex]V=-4.01\times 10^4\ volts[/tex]
Explanation:
Given that,
Charge 1, [tex]q_1=2.2\ \mu C[/tex]
Point charge 2, [tex]q_2=-8\ \mu C[/tex]
Distance between charges, d = 2.6 m
We need to find the electric potential midway between them. The electric potential is given by :
[tex]V=\dfrac{kq}{r}[/tex]
In this case, r = 1.3 m (midway between charges)
[tex]V=\dfrac{kq_1}{r}-\dfrac{kq_2}{r}[/tex]
[tex]V=\dfrac{k}{r}(q_1-q_2)[/tex]
[tex]V=\dfrac{9\times 10^9}{1.3}(2.2\times 10^{-6}-8\times 10^{-6})[/tex]
[tex]V=-40153.84\ volts[/tex]
or
[tex]V=-4.01\times 10^4\ volts[/tex]
So, the electric potential midway between the charges is [tex]V=-4.01\times 10^4\ volts[/tex]. Hence, this is the required solution.
