Free body diagram
Magnitude and net charge of q2
Please..
Solution :
Given :
q1 = 3 µC = 3 x 10-6 C
q2 = - 4 µC = - 4 x 10-6 C
q3 = - 7 µC = - 7 x 10-6 C
.
.
Here, Force on Charge q2 due to q1 will be towards right and Force on Charge q2 due to q3 will be towards Left.
Where,
And,
Therefore :
Magnitude of the net force on the charge q2 will be :
Fnet = F12 - F32 = 2.7 N - 2.057 N = 0.643 N (Towards Right)
Free body diagram Magnitude and net charge of q2 Please.. 0.20 m O.15 m 42 41...
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