A +2.50C charge is located at a point A at the origin, and a -6.00 C charge is l

Question

A +2.50C charge is located at a point A at the origin, and a -6.00 C charge is located at point B at +20.0cm on the x-axis. Point C is located at +15.0 cm on the y-axis.

A) Find the magnitude of the net electric field at point C.

B) Find the magnitude of the net force due to the electric field on a 2.0 C charge located at point C.

Explanation / Answer

First find the field due to A, E=kQ/r^2 = (9 x 10^9)(2.5 x 10^-6)/(.15)^2 = 1000000 N/C pointing up, since repulsive force along y axis Field due to B, E=kQ/r^2 But we need the distance from B to C, so using pythagorean theorem, (20)^2 + (15)^2 = x^2 x= 25 cm = .25 m We should also find the angle, we will use it later tan x = 15/20 x= 36.87 degrees E=kQ/r^2 = (9 x 10^9)(6 x 10^-6) / (.25)^2 = 864000 N/C Since this is attractive force, the x component is positive, and y component is negative cos 36.87 = x/ 864000 x= 691199 N/C is x component sin 36.87 = y/864000 y= -518401 N/C is y component Now add up all x components ==> 691199 N/C y components==> 1000000 - 518401 = 481599 N/C Now to find magnitude, just use pythagorean theorem = 842433 N/C B) Now F=Eq, so just multiply the E by 2 x 10^-6 C F= (842433)(2 x 10^-6) = 1.685 N

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