Electric field at a point (point charge and semi-circle rod)

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SUMMARY

The discussion focuses on the electric field at point C due to a negative point charge -Q and a uniformly charged semicircular rod with total charge +Q. It concludes that the magnitude of the electric field due to the point charge (Ep) is greater than that of the rod (Er) because the rod can be treated as a series of point charges contributing to the net electric field. The net electric field at point C is equal to Ep, as the electric field lines from the rod converge at that point, reinforcing the field created by the point charge.

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AnonBae
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Homework Statement


A thin semicircular rod has a total charge +Q uniformly distributed along it. A negative point charge -Q is placed as shown. (Point C is equidistant from -Q and from all points on the rod.)
*Image Attached*
Let Ep and Er represent the electric fields at point C due to the point charge and the rod respectively.

a) Is the magnitude of Ep greater than, less than, or equal to the magnitude of Er? Explain.

b) Is the magnitude of the net electric field at point C greater than, lessthan, or equal to the magnitude of Ep? Explain.

Homework Equations


ΣE = Fe / q

The Attempt at a Solution


a. I am guessing the magnitude of Ep is greater than the magnitude of Er since we can treat a segment charge on the rod as a point charge, and thus the by the equation, the electric field is dependent on the charge. The fractional charge on the rod will produce a greater electrical force than the point charge does.
b. The magnitude of the net electric field is equal to the magnitude at Ep since the point C lies on the electric field line between point C and the fractional charge on the rod.
 

Attachments

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AnonBae said:
we can treat a segment charge on the rod as a point charge
In what sense, placed where, and of what magnitude?
AnonBae said:
The fractional charge on the rod
Why fractional?

For b, I don't think you are reading the question correctly.
 
a. I do know that a fractional charge of +Q contributes to the electric field at point C since no other electric field lines cross through point C besides the one horizontal to point C. I am confused as to whether that fractional charge offsets its weak electric force by the formula for the electric field or it does not.

b. So, the net electric field is contributed by all charges on the rod (total of +Q) and the sum of the electric field from all those charges will equal the electric field Ep since they both converge at the same point?
 
AnonBae said:
fractional charge of +Q contributes to the electric field at point C
I still don't know what you mean by a fractional charge here. Do you mean a continuously distributed charge? Or one infinitesimal element of the distributed charge? Or just some fraction less than 1 of Q?
AnonBae said:
no other electric field lines cross through point C besides the one horizontal to point C.
Except for the extreme ends, all parts of the rod generate a field which has a horizontal component at C.
Write an integral for the total field the rod produces at C.
 

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