each would create a E vector pointing through the center from the charge segment. The components perpendicular to the diameter add. The components parallel to the diameter cancel each other. So you end up with an E vector perpendicular to that diameter.
If you do the other half as well:
NO look at it. If you complete the circle, those E vectors are equal and opposite :) NO field at the center by symmetry.
Consider a wire in the shape of a semi-circle of radius R, with a total positive net charge Q distributed over the
wire uniformly. Divide the wire into at least 8 equal point-like “chunks” of charge. Then, for each “chunk,”
sketch (on the same diagram) the E-field vector that bit of charge would cause at the center of the (semi-) circle.
Explain, using this picture, how each “chunk’s” field contributes to the net field at that center point.
Would a full circle of charge double the net field at the center? Explain.
2 answers
What do you mean when you say NO?
1 answer