Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m (see (Figure 1) ). NEED HELP!!!!! [Solved]

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

Get Answer for  a quanti ml corrispondono 20 cc [Solved]
Part F Calculate the direction of the net force on particle 3 due to the other two. tan = Fy/Fx = -0.2031/-0.2153 = arctan -2.5786 = -43.33 from -y-axis (since its on Quadrant III)

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