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

(I have been given 5 separate answer and webassign has said that they are all incorrect)

Explanation / Answer

the distance between + 40 nc to 100 nc is

from the pythogerian theorem

r = root 9^2 + 10 ^2 = 13.45 cm = 0.1345 m

tan theta = 9/10

theta = tan^-1 ( 9/10 ) = 41.98

from the Coloumb law the force between + 40 nc to 100 nc is

F1 = kq1 q2/r^2 = ( 9* 10 ^9) ( 40 * 10 ^-9 C) ( 100 * 10 ^-9 C) /( 0.1345 m)^2 ( cos 41.98 i + sin 41.98 j)

                    =1.479 * 10 ^-3 i +1.33 * 10 ^-3 j

force due to - 40 nc is

F2 = kq1 q2/r^2 = ( 9* 10 ^9) ( -40 * 10 ^-9 C) ( 100 * 10 ^-9 C) /( 9 * 10 ^-2 m)^2 =-4.44 * 10 ^-3 j

the net force is

F = F1 + F2 = 1.479 * 10 ^-3 i +1.33 * 10 ^-3 j-4.44 * 10 ^-3 j = 1.479 * 10 ^-3 i-3.11 * 10 ^-3 j

magnitdue of force is

F = root( 1.479 * 10 ^-3 )^2+(3.11 * 10 ^-3)^2 = 3.44* 10 ^-3 N

the electric field at the location of the +100.0nC charge due to the presence of the other two charges is

E = F/q = 3.44* 10 ^-3 N/100 * 10 ^-9 C = 34400 N/C

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