Two point charges, +2.80 μC and -6.20 μC, are separated by 2.30 m. What is the electric potential midway between them?

V=-26651.7V

Explanation:

The electric potential is for an electric charge is expressed as

V=kq/r.

Where K is constant, q is the charge and r is the distance.

Hence for a charge of +2.80 μC and - 6.20μC, both at distance 2.3m to each other, we can calculate the electric potential midway between them as

V=(kq/r)-(kq/r)

Since the mid way distance is 2.3/2=1.15m

Hence

V=K[(q/1.15) - (q/1.15)]

V=9*10⁹*10⁻⁶(2.43-5.39)

V=-26651.7V

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