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.

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.

What do you mean when you say NO?

To determine the E-field vector at the center of the semi-circle, we need to consider the contribution from each chunk of charge along the wire. We can calculate the E-field using Coulomb's Law, which states that the magnitude of the electric field created by a point charge is directly proportional to the charge and inversely proportional to the square of the distance.

To sketch the E-field vector for each chunk, we can use the principle that the field lines emanate radially from a point charge. At each chunk, draw an arrow in the direction of the electric field, with the length of the arrow representing the magnitude of the field.

Now, let's consider how each chunk's field contributes to the net field at the center point:

1. Chunks on the vertical diameter: The electric fields from these chunks point directly toward or away from the center. At the center, these fields will add up because they have the same direction. Therefore, the net field from the chunks on the vertical diameter will have a vertical component.

2. Chunks on the semi-circular part: The electric fields from these chunks will have both horizontal and vertical components. However, since the horizontal components from opposite sides cancel out due to symmetry, only the vertical components will contribute to the net field at the center.

3. Chunks on the upper semi-circle: The vertical components of the electric field from these chunks will add up with the vertical components from the chunks on the vertical diameter, increasing the net field at the center.

4. Chunks on the lower semi-circle: The vertical components of the electric field from these chunks will partially cancel out with the vertical components from the chunks on the vertical diameter, decreasing the net field at the center.

By considering the contributions from each chunk, we can determine the net field at the center of the semi-circle.

Now let's address the second part of the question. If we were to form a full circle of charge with the same net positive charge Q and divide it into similar chunks, the net field at the center would be the same as before. This is because the vertical components from the chunks on the upper semi-circle still add up, while the vertical components from the chunks on the lower semi-circle still partially cancel out. Therefore, the net field at the center does not double with a full circle of charge.

In summary, when dividing the wire into chunks of charge and considering their contributions, each chunk's field adds up to create the net field at the center. The net field can be determined by adding the vertical components from the chunks on the vertical diameter and the upper semi-circle while subtracting the vertical components from the chunks on the lower semi-circle. A full circle of charge does not double the net field at the center because the vertical components still follow the same pattern as in the semi-circle case.