Homework Help: Vector Velocity and Coulomb's Law

1. May 9, 2010

sami23

1. The problem statement, all variables and given/known data
Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.5 cm. Two of the particles have a negative charge: q_1 = -8.0 nC and q_2 = -16.0 nC. The remaining particle has a positive charge, q_3 = 8.0 nC. Assume that particle 3 is no longer fixed to a corner of the triangle and is now allowed to move. In what direction would particle 3 move the instant after being released???

**How do I draw the velocity vector for particle 3

2. Relevant equations
Coulomb's Law: F = [k*q1*q2]/r2

3. The attempt at a solution
F13 (on q3 due to q1) = [k *q3*q1]/r2

= [(9*109)(8*10-9)(8*10-9)]/(.035)2

= 4.702*10-4 N

F23 (on q3 due to q2) = [k * q3*q2]/ r2

= [(9*109)(8*10-9)(16*10-9)]/(.035)2

= 9.404*10-4 N

F13x = F13 * cos(60) = + 2.351*10^-4 N
F13y = F13 * sin(60) = + 4.072*10^-4 N
F23x = F23 * cos(0) = + 9.404 *10^-4 N
F23y = 0

Fx = 2.351*10^-4 + 9.404*10^-4 = 1.176*10^-3 N
Fy = 4.072*10^-4 N

ΣF = 10^-3√(1.176)2 + (0.4072)2 = 1.2445*10^-3 N
θ = arctan (0.4072/1.176) = 19.1

The velocity of the body will be directed with the force as the body was initially at rest
F gives the direction of acceleration a
a = (change in velocity ) /time
a=(v-0)/t
a is directed towards F but a, v, and t are unknowns

2. May 9, 2010

diazona

Yes. Now do you know the direction of the force? (Hint: yes you do)

3. May 9, 2010

sami23

so it would go in the (+x axis, +y axis) with θ = 19?