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.

(a) In terms of the symbolic quantities given above and on the diagram, what is the charge per unit length of the rod?
=

(b) What is the amount of charge dQ on the small piece of length dx?
dQ =

(d) What is the distance from the source to the observation location?
d =

(e) When we set up an integral to find the electric field at the observation location due to the entire rod, what will be the integration variable?

Explanation / Answer

a) lamda = charge on rod / length of the rod-Q/2A

b) lamda = dQ/dx = (-Q/2A)

dQ = (-Q/2A)*dx

c) r = -x i + y j

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

e) x

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