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

Answer 1

as we know, charges with same sign repel each other and charges with opposite sign attract each other.
force between two charges q1 and q2, separated by a distance of d is given by

k*q1*q2/d^2

where k=9*10^9

a)force on 100 nC charge due to +40 nC charge:

as both the charges are positive, force is repulsive in nature.

distance=sqrt(10^2+9^2)=13.453 cm

force magnitude=9*10^9*40*10^(-9)*100*10^(-9)/(0.13453^2)=1.989 mN

hence it will be in the direction from +40 nC to +100 nC .

vector along this direction=(10,9)-(0,0)=10 i+ 9 j

unit vector along this direction= (10,9)/sqrt(10^2+9^2)=(0.743,0.67)

in vector notation, force =1.989*(0.643,0.67)=(1.278,1.332) mN

b)force on 100 nC charge due to -40 nC charge:

as the charges are of opposite nature, force is attractive in nature.

distance=9 cm

force magnitude=9*10^9*40*10^(-9)*100*10^(-9)/(0.09^2)=4.45 mN

it will be in the direction from +100 nC to -40 nC .

vector along this direction=(10,0)-(10,9)=- 9 j

unit vector along this direction= -j

in vector notation, force =-4.45 j mN

hence net force=(1.278 i -3.118 j )mN

B) electric field=force/charge=(1.278 i -3.118 j ) * 10^4 N/C