Consider a wire in the shape of a semi-circle of radius R, with a total positive net charge Q vistributed 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.
I cant do the sketch. Notice the E is in directions, which the vertical will cancel out, and you are left with a net horzontal (to the axis of symettry) component.
Given that, a full circle would then cancel that out, and E at the center would be zero.
Now, all that is true in the Plane of the semicircle. However, above and below the that plane, it is not true, there will be a component in the z direction on each side of the plane. Think about that (pretend instead of the wire, you had a disk of charge)...
To understand how each "chunk's" field contributes to the net field at the center point, let's break it down step by step using a diagram.
1. Divide the wire into at least 8 equal point-like "chunks" of charge. Each chunk represents a small portion of the total charge Q.
2. At the center of the semi-circle, draw a point as the reference point. This point represents the location where we want to determine the net electric field.
3. Now, consider each chunk of charge individually. For simplicity, let's start with the first chunk of charge located at the bottom of the semi-circle.
4. Since the chunk of charge is distributed uniformly along the wire, its contribution to the electric field at the center point can be considered as if the entire charge is concentrated at the position of the chunk.
5. To find the electric field produced by this chunk at the center, you can use Coulomb's Law. The electric field vector (E-field) points radially outward from the chunk of charge, with a magnitude given by:
E = k * (q / r^2)
where k is the electrostatic constant (k = 9 x 10^9 Nm^2/C^2), q is the magnitude of the charge of the chunk, and r is the distance between the chunk and the center point.
6. Repeat the process for each chunk of charge, determining the electric field at the center for each chunk using Coulomb's Law.
7. Now, to determine the net electric field at the center, add up the contributions of the individual electric field vectors produced by each chunk of charge. Since the electric field is a vector quantity, you need to consider both the magnitude and direction of each contributing field vector. By adding them up vectorially, you can calculate the net field at the center.
8. Once you have determined the net electric field at the center point, you can sketch it on the same diagram as vectors originating from the center point. The resultant vector should represent the sum of all the individual electric field vectors at that point.
Regarding a full circle of charge doubling the net field at the center, the answer is no. If you were to distribute the charge uniformly over the entire circle, the net electric field at the center would be the same as that of the semi-circle. This is because the contributions from opposite sides of the circle would cancel each other out. The individual electric field vectors would have equal magnitudes but opposite directions, resulting in a net effect of zero.