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-Q and from all points on the rod.) Let F rightarrow p, and F rightarrow R, represent the force on the test charge due to the point charge and the rod respectively. Is the magnitude of F rightarrow p greater than, less than, or equal to the magnitude of F rightarrow R? Explain how you can tell. Is the magnitude of the net force on +q greater than, less than, or equal to the magnitude of F rightarrow p? 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

Force from -Q on +Q, . the force Fp , is entirely in one direction.

The force from the charge on the rod, however, is in manydirections. All of them have some component of force to the left(repulsive force) but there is also an up or down component to theforce from each bit of charge on the rod.

Therefore, some of the force from the charge on the rodcancels out. For this reason:   Fp   is greaterthan Fr.

(b)   The force from -Q  on   + Q is to the left (attractive force). The force from the rod on +Q is also to the left (repulsive).

Therefore net force = Fp + Fr        since they are in the same direction.

And therefore net force is greaterthan Fp.

(c)

when the second negetive charge is not there the force on the +q charge is to the left and is equal to Fp+Fr

but when u place the second negetive charge, the force due tothis charge is to right , let this be Fs then the netforce is = Fp + Fr - Fs

so the magnitude of the net electric force on +q less than the magnitude of the net electric force on +q inpart b.

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