The 69 nC particle is between the other two point charges. One exerts a push one it and the other exerts a pull, but since they are on opposite sides, the two force add.
Use Coulomb's law (twice) to get the net force.
Divide the net electrostatic force by the acceleration to get the mass.
If following this procedure does not get you the right answer, please show your work so someone can further assist you.
Include the units of the mass with your answer, which should be kg. Numbers alone, without dimensions, are of no value and deserve to be marked wrong
Two point charges of 48 nC and -45 nC are held fixed on an x axis, at the origin and at x = 77 cm, respectively. A particle with a charge of 69 μC is released from rest at x = 49 cm. If the initial acceleration of the particle has a magnitude of 123 km/s2, what is the particle's mass?
I got a value of 4.033E-6 but i cannot figure out where my mistake is since i double checked my work with a friend who got the correct answer
3 answers
F13=k*48E-9*69E-6/.28^2 which i got to be .37978 N
F23=K845E-9*69E-6/.49^2 which i got to be .11625N
Ftotal=.49603N
M=F/a so .49603/123000m/s which i got to be 4.0327kg
F23=K845E-9*69E-6/.49^2 which i got to be .11625N
Ftotal=.49603N
M=F/a so .49603/123000m/s which i got to be 4.0327kg
The 48 nC particle is 0.49 m away, not 0.28 m.
The -45 nC particle is 0.29 m away.
You may have got them mixed up.
The -45 nC particle is 0.29 m away.
You may have got them mixed up.
1 answer