The charges are Q1 = 6.7 C , Q2 = -9.5 C , and Q3 = -5.6 C .

Ok so here goes:

First determine the force acting between the charges 2 at a time using the formula:

F = k*q1*q2/d^2 where k =910^9 Nm^2/C^2

F12 = (910^9)*(6.7 )*(9.5 )/(1.2^2) = 0.3978 N (we disregard the minus sign)

F23 = (910^9)*(9.5 )*(5.6 )/(1.2^2) = 0.3325 N

F13 = (910^9)*(6.7 )*(5.6 )/(1.2^2) = 0.2345 N

Part A

Calculate the magnitude of the net force on particle 1 due to the other two.

Q1 attracts both Q2 and Q3 so that:

Fx = F12cos60 F13cos60 = 0.3978 cos60 0.2345cos60

Fx = 0.08165 N

Fy = F12sin60 F13sin60 = 0.3978 sin60 0.2345sin60

Fy = 0.5476 N

Resultant (R1) = sqrt(Fx^2 Fy^2) = sqrt(0.08165^2 0.5476^2)

R1 = 0.5537 N <--ANSWER
Part B
Calculate the direction of the net force on particle 1 due to the other two.
tan = Fy/Fx = 0.5476/0.08165
= arctan 6.7066
= 81.52 degrees from x-axis <---ANSWER
Part C
Calculate the magnitude of the net force on particle 2 due to the other two.
Q2 attracts Q1 and repels Q3
Fx = - F12cos60 F23 = -0.3978 cos60 0.3325
Fx = 0.1336 N
Fy = - F12sin60 0 = -0.3978sin60 (zero was used because F23 has no vertical component)
Fy = -0.3445 N
Resultant (R2) = sqrt(Fx^2 Fy^2) = sqrt(0.1336^2 (-0.3445)^2)
R2 = 0.3695 N <--ANSWER
Part D
Calculate the direction of the net force on particle 2 due to the other two.
tan = Fy/Fx = -0.3445/0.1336
= arctan -2.5786
= -68.80 degrees from x-axis <---ANSWER
Part E
Calculate the magnitude of the net force on particle 3 due to the other two.
Q23 attracts Q1 and repels Q2
Fx = F13cos60 - F23 = 0.2345 cos60 - 0.3325
Fx = -0.2153 N
Fy = - F13sin60 0 = -0.2345 sin60 (zero was used because F23 has no vertical component)
Fy = -0.2031 N
Resultant (R2) = sqrt(Fx^2 Fy^2) = sqrt((-0.2153)^2 (-0.2031)^2)
R2 = 0.296 N <--ANSWER

its a triangle with 1.2m from each other and the two on the bottom are positive and one on top is negative

This post is last updated on hrtanswers.com at Date : 1st of September – 2022