Three point charges are placed on the x-y plane: a +40.0nC charge at the origin,

Question

Three point charges are placed on the x-y plane: a +40.0nC charge at the origin, a -40.0nC charge on the x axis at 10.0cm, and a +100.0nC charge at the point (10.0cm, 9.00cm).

(A) Find the total electric force on the +100.0nC charge due to the other two.

Please answer in this format: ( ? N)x + ( ? N)y

(B) What is the electric field at the location of the +100.0nC charge due to the presence of the other two charges?

Please answer in this format: ( ? N/C)x + ( ? N/C)y

Someone answered with:

electric field = (14783.1 N/C)x - (31138.3 N/C)y

force = (0.00147831 N)x - (0.00311383 N)y

Can someone check and confirm or deny if this answer is correct?

Explanation / Answer

Electric field vector due to +40 nC at the point (10 cm, 9 cm) will make an angle = tan-1(9/10) = 41.987o from the x-axis.

So, electric field at (10 cm, 9 cm) due to +40 nC is:

E1 = kQ/r2 = [(8.99 * 109)(40 * 10-9)/(0.12 + 0.092)] [(cos 41.987o) x + (sin 41.987o) y] N = (14767.4 N) x + (13290.5 N) y

Electric field due to -40 nC at the point (10 cm, 9 cm) is:

E2 = (8.99 * 109)(-40 * 10-9)/0.092 y =  (-44395.1 N) y

So, net electric field at (10 cm, 9 cm) is:

Enet = E1 + E2 = (14767.4 N)x + (13290.5 N) y - (44395.1 N) y = (14767.4 N) x - (31104.6 N) y

Net electric force on 100 nC charge is, Fnet = (100 nC) Enet

=> Fnet = (100 * 10-9)((14767.4 N) x - (31104.6 N) y) = (0.0014767 N) x - (0.00311046 N) y

The difference in answers is due to the value of Coulumb's constant used; I used its value as 8.99 * 109 as opposed to the less accurate 9 * 109 in the answer you mentioned.

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