Coulombs law, and 3 point particles (vector question)

[michaelw]

Coulombs law, and 3 "point particles" (vector question)

The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q1=+8.00 mC; the other two have identical magnitudes, but opposite signs: q2=–5.00 mC and q3=+5.00 mC. (a) Determine the net force (magnitude and direction) exerted on q1 by the other two charges. (b) If q1 had a mass of 1.50 g and it were free to move, what would be its acceleration?


for some reason i think i did this wrong

abs(F13) = abs(F12)
= (8.99*10^9)(8*10^-6)(5*10^-6)/(1.3^2)
= 2.13*10^-19

the F13 will make q1 go up and left, F12 will make q1 go up and right.
the left and the right forces cancel, meaning that the q1 will travel directly up? (by a magnitude of 2*sin(23)*F12 => 1.66*10^-19)

 

[MathStudent]

The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q1=+8.00 mC; the other two have identical magnitudes, but opposite signs: q2=–5.00 mC and q3=+5.00 mC. (a) Determine the net force (magnitude and direction) exerted on q1 by the other two charges. (b) If q1 had a mass of 1.50 g and it were free to move, what would be its acceleration?


for some reason i think i did this wrong

abs(F13) = abs(F12)
= (8.99*10^9)(8*10^-6)(5*10^-6)/(1.3^2)
= 2.13*10^-19

The question gives the charges in mC which is milli coulombs (10^-3 C), you used micro coulombs
($\mu C =$ 10^-6 C.)
argh...your like the third person today to make the same mistake

any way, even if it were in micro coulombs, the equation is right, but the calculation is incorrect... how did you get 10^-19??

the F13 will make q1 go up and left, F12 will make q1 go up and right.
the left and the right forces cancel, meaning that the q1 will travel directly up? (by a magnitude of 2*sin(23)*F12 => 1.66*10^-19)

right!

 

[wais]

Your calculations for the magnitude of the force between q1 and q3 (F13) and the direction of the net force on q1 are correct. However, in order to calculate the acceleration of q1, we need to use Newton's second law, which states that F=ma. This means that the acceleration of q1 will be equal to the net force on q1 (which you have correctly calculated) divided by its mass (given as 1.50 g). So, the acceleration of q1 would be 1.66*10^-19/0.0015 = 1.11*10^-16 m/s^2. Keep in mind that this is a very small acceleration due to the small masses and distances involved.