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

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14