- #1
michaelw
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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)
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)