A thin semicircular rod has a total charge +Q uniformly distributed along it. A

Question

A thin semicircular rod has a total charge +Q uniformly distributed along it. A negative point charge - Q is placed as shown. A test charge +q is placed at point C (point C is equidistant from all points on the rod.). Let F_P and F_R represent the force on the test . charge and the rod respectively. Is the magnitude of the net force on +q than, less than, or equal to the magnitude of Explain. A second negative point charge -Q is placed as shown. Is the magnitude of the net electric force on +q greater than, less than, or equal to the magnitude of the net electric force on +q in part b? Explain.

Explanation / Answer

a) The magnitude of the force on the test charge by the point charge (FP) greater than the magnitude of the force on the test charge by the rod (FR), because the y-components of segments of the rod cancelling each other out.

b) The charge at point "C" is positive, so it is being attracted to the negative point charge. It is also being repelled from the positively charged rod, so FP and FR are in the same direction, and probably add, given the symmetry of the configuration. If they add, the net force is greater than FP, since it includes FP.

c) Now second charge -Q will attract the +q charge, that force will be in opposite direction than the direction in part (b).So the net force on +q will be lesser than force in part (b).

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