The above Figure shows the rod. Highlighted is one small piece of the rod which

Question

The above Figure shows the rod. Highlighted is one small piece of the rod which is located at a position x and has length (dx). and upper limits on the integration? Answer in terms of symbols x, L, D, (dx) and constants. For a finite continuous charged object we can use integration to find the electric potential at a point. As is the case for any finite charge distribution, the value of the electric potential can be chosen to be zero at any reference point which is sufficiently far from the distribution. I.e. One can (and usually does) choose Consider a rod of length L extending from the origin along the positive x axis, and a point P located at x = -D, which is a distance D from the origin. The task is to find the electric potential (voltage) at the point P. The above Figure shows the rod. Highlighted is one small piece of the rod which is located at a position x and has length (dx). (a) Suppose that the rod is uniformly charged to a value Q. Write the expression for the linear charge density, using the symbols already defined. lambda = ? (b) The small piece of the rod has length dx. In terms of symbols L, x, (dx), D, Q and constants, what is the infinitesimal charge on the rod? dq = ? c) What is the distance from the point to the small piece of the rod? Answer in terms of symbols x, L, D, (dx) and constants. (d) What is the infinitesimal voltage produced at point P from this dq? Answer in terms of symbols k, Q, x, L, D, (dx) and constants. dV = ? (e) The final step would be to integrate this result. What are the lower and upper limits on the integration? Answer in terms of symbols x, L, D, (dx) and constants. Note that if the charge of the rod had not been uniform, we would need to know lambda as a function of x and use that in place of (Q/L).

Explanation / Answer

a.) Q/L b.) (Q/L)(dx) c.) D + x d.) (k(Q/L)(dx))/(D+x) e.) lower limit = 0, upper limit = L

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