A thin rod lies on the x -axis with one end at - A and the other end at A , as s

Question

A thin rod lies on the x-axis with one end at -A and the other end at A, as shown in the diagram. A charge of -Q is spread uniformly over the surface of the rod. We want to set up an integral to find the electric field at location ‹ 0, y, 0 › due to the rod. Following the procedure discussed in the textbook, we have cut up the rod into small segments, each of which can be considered as a point charge. We have selected a typical piece, shown in red on the diagram.

Use the following as necessary: x, y, dx, A, Q. Remember that the rod has charge -Q.

Explanation / Answer

a) lambda = charge / length = -Q / 2A

b) dQ = dx * lambda = - Qdx / 2A

c) r = (x, 0,0) - (0, y , 0)

r = <x, -y, 0>

d) d = sqrt(x^2 + y^2)

e) intergration variable will be x.

y is fixed and constant.

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