Two charges are fixed in an xy coordinate system Charge 1 3

Two charges are fixed in an x-y coordinate system. Charge 1, -3 uC is located in the origin. Charge 2, 8.0 uC is located on the y-axis at x = 0.0 m, y = 0.4 m. Find the electric potential at a point A on the x-axis, located at x = 0.30 m, y = 0.0 m. (k = 8.99 * 10^9 N m^2/C^2)

Solution

the electric potential V due to the charge particle is V= k(q)/r,

and hence,

V= kq₁/r₁+kq₂/r₂

Where r1 and r2 is the distance between each of the charges from a pont A,

r1= 0.30 m and r2= sqrt((0.30²+0.4²)

therefore,

V= k(-3*10^-6)/0.30+k(8*10^-6)/0.5

=k(-10*10^-6+ 16*10^-6)

=k(6*10^-6)= 8.99*10⁹*6*10^-6

V= 53.94* 10³ V