1. If two equal charges are separated by a certain distance, the force of repuls

Question

1.       If two equal charges are separated by a certain distance, the force of repulsion is F. Given F1, if each charge in F1 is doubled and the distance is cut in half, what is the new value of the force of repulsion, F2? Express your answer in terms of F1.<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

2.       The electrical force on a 2-coulomb charge is 60 Newtons (N). What is the magnitude of the electric field at the place where the charge is located?

3.       What is an electric field? How do you determine the magnitude and direction of the field at a given point?

Explanation / Answer

1) F2 = 8(F1) Use the formula F = kq1q2/(r^2): if force is doubled, then new force will be doubled. however; if distance is cut in half, then r^2 is cut in quarter and since it's in denominator, it will be quadraple. so 2x4=8 2) E = F/q = 60N/2C = 30N/C 3) Electric field is caused by force acting on each charges and direction may change due to movement of charge.

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