The net force on the charge at the origin is -1.2×10-8
<u>Explanation:</u>
Solving the problem,
- Draw the x-axis and the locations of the given three charges.
- The forces applied on the charge at the origin and there are two of them, and since all the changes are positive, all the forces are repulsive.
- we have the formula, F = kq1Q/r².
- F1 = kq1Q/r²1 = (9.0*109Nm²/C²)(2.2*10^-9C)(3.5*10^-9C)/(1.5m)² = 31*10-9N = 3.1*10-8N. F1 points to the right (+x direction).
- F2 = kq2Q/r²2 = (9.0*109Nm²/C²)(5.4*10^-9C)(3.5*10^-9C)/(2.0m)² = 43*10^-9N = 4.3*10^-8N.
- F2 points to the left (-x direction).
- To find the net force we have to subtract the force F1 and force F2 .
- The net force is F(origin) = F1 - F2 = -1.2×10-8N.
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It would not. Imagine four forces equal in magnitude but opposite in direction (e.g. north, east, south, and west). If these forces were to double in magnitude they would still have the same magnitude, meaning the net force is still equal to zero.
Answer:
since -6 lasted for 5 seconds, multiplying both would result in -30
3 lasted for 10 seconds, so multiplying both would give +30
average = ( 30 + (-30) ) / 2
30 -30 is already equal to zero, so the answer should be 0
A coil of insulated wire around an iron core
