Question

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 2.1 cm . Two of the particles have a negative charge: q1 = -8.0 nC and q2 = -16.0 nC . The remaining particle has a positive charge, q3 = 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?

A) Find the net force ΣF⃗ 3 acting on particle 3 due to the presence of the other two particles. Report you answer as a magnitude ΣF3 and a direction θ measured from the positive x axis.

Please show the process! I've tried everything I can think of.

Answer 1

step 1:

lets assign coordinates to the corners.

let the bottom two corners are at q1 at (0,0) and q2 at(a,0) where a=2.1 cm

then x coordinate of third corner=a/2 (because perpendicular from one corner to the line joining other two corners bisects the line)

y coordinate=sqrt(3)*a/2 (height of the equilateral triangle=sqrt(3)*length of a side/2)

hence its coordinates are (0.5*a,0.866*a)

now, force on particle 3 due to q1:

as q1 and q3 are of opposite nature, the force will be attractive in anture.

hence it will be away from q3 and towards q1.

direction of force in vector notation=(0,0)-(0.5*a,0.866*a)=(-0.5*a,-0.866*a)

distance=a

so force magnitude=9*10^9*q1*q3/a^2=1.306*10^(-3) N

in vector notation, force=(-0.653,-1.131)*10^(-3) N

now, force on particle 3 due to q2:

as q1 and q3 are of opposite nature, the force will be attractive in anture.

hence it will be away from q3 and towards q2.

direction of force in vector notation=(a,0)-(0.5*a,0.866*a)=(0.5*a,-0.866*a)

distance=a

so force magnitude=9*10^9*q2*q3/a^2=2.6122*10^(-3) N

in vector notation, force=(1.3061,-2.2622)*10^(-3) N

hence net force=(-0.653,-1.131)*10^(-3)+ (1.3061,-2.2622)*10^(-3)=(0.65312,-3.3932)*10^(-3) N

magnitude of force=sqrt(0.65312^2+3.3932^2)*10^(-3)=3.4555*10^(-3) N

direction with +ve x axis=arctan(-3.3932/0.65312)=-79.105 degrees