A positive charge with twice the value of the intial charge (on the left of the
ID: 1260210 • Letter: A
Question
A positive charge with twice the value of the intial charge (on the left of the loop) is place to the right of a loop. Both charges are the same distance from the loop and placed along the axis of the loop.
1. Is the net electric flux throught he loop due to the charge postivie, negative, or zero? Explain reasoning
2. Suppose the new charge located on the right of the loop had been negative instead of positive. How would your answer to part number 1 change if at all? Explain.
Explanation / Answer
Electric Flux = E (dot) Area = EA cos (theta)
a) I drew the area vector perpendicular to the plane of the circular loop, pointing to the left. And then I said the electric field lines are coming out of +Q and passing through the loop. So E (dot) A would be positive, because Ex and A are both pointing in the same direction ( to the left), and the Ey components all cancel out.
b) I said in this case, there are more Ex lines from the +2Q charge becuase it has a great magnitude, so the net Ex is to the right (because the +2Q charges electric field lines point to the right). Since I had made my area vector point to the left, the directions of the Area vector and the net Ex are in opposite directions, so the net electric flux through the loop is negative.
c) I said that the Ex due to -2Q would have direction to the left, because it is negative, so the net Ex would be to the left. And since my area vector is to the left, and the net electric field Ex is to the left, the electric flux would have been positive not negative had it been -2Q
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