Question

A charge q₁=4.10 mCis located at the origin of a coordinate system. Another charge q₂=4.80 mC is located at x=-1.70 m and y=-4.50 m, and another q₃=4.20 mC is located at x=-2.30 m and y=1.20 m. Calculate the magnitude of the net electric force on the charge q₁ in Newtons (N).

Answer 1

electric force between similar signed charges is repulsive in nature and force between opposite charges is attractive in nature.

force on q1 due to q2:

as q1 and q2 are of same sign, force is repulsive in nature and directed from q2 towards q1.

direction of force=(0,0)-(-1.7,-4.5)=(1.7,4.5)

distance=sqrt(1.7^2+4.5^2)=4.81 m

unit vector along force=(1.7,4.5)/4.81

force magnitude=k*q1*q2/distance^2

where k=coloumb’s constant=9*10^9

force magnitude=9*10^9*4.1*10^(-3)*4.8*10^(-3)/4.81^2

=7655.6 N

in vector notation, force on q1 due to q2=F1=7655.6*(0.35343,0.93555) N

force on q1 due to q3:

as q1 and q3 are of same sign, force is repulsive in nature and directed from q2 towards q1.

direction of force=(0,0)-(-2.3,1.2)=(2.3,-1.2)

distance=sqrt(2.3^2+1.2^2)=2.5942 m

unit vector along force=(2.3,-1.2)/2.5942

force magnitude=k*q1*q3/distance^2

where k=coloumb’s constant=9*10^9

force magnitude=9*10^9*4.1*10^(-3)*4.2*10^(-3)/2.5942^2

=2.3029*10^4 N

in vector notation, force on q1 due to q3=F2=2.3029*10^4*(0.8866,-0.46257) N

net force=F1+F2=(2.3123,-0.349)*10^4 N

magnitude=sqrt(2.3123^2+0.349^2)*10^4 N

=2.3385*10^4 N