Question

1.(coulombs law)if a=3.0mm,b=4.0mm, Q1=60nC,Q2=-80nC, and q=36nC, what is the magnitude of the electric force on q?
2.)A particle (m=50g,q=5.0mC)is released from rest when it is 50 cm from a second particle (Q=-20mC).determine the magnitude of the initial accelertion?

Answer 1

For the part 2), the following free body diagram is drawn. Remember that the two charges will attract each other

The electrostatic force is given by\[Fe = kqQ/r ^{2}\]
r = 50 cm
q and Q are given.

Force due to gravity
\[Fg = mg\]
The total downward force =\[F = Fe+Fg\]
For acceleration we use Newton's law\[a = F /m =(Fe + Fg)/m = kqQ/r ^{2} m + g\]
Acceleration due to electrostatic force will probably be much greater than g and hence the latter can be neglected.

For your first part please provide the figure.

Answer 2

Another important thing.
The q and Q are magnitudes of charges. While finding Fe put only the magnitudes of q and Q. The values of Fe and Fg are positive as we have already recognized the direction to be vertically downwards.

Answer 3

thanks