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23 Nov 2019

For a finite continuous charged object we can use integration tofind the electric potential at a point. As is the case for anyfinite charge distribution, the value of the electric potential canbe chosen to be zero at any reference point which is sufficientlyfar from the distribution. I.e. One can (and usually does) chooseV_(infinity) = 0 at r = infinity.

Consider a rod of length L extending from the origin along thepositive x axis, and a point P located at x = -D, which is adistance D from the origin. The task is to find the electricpotential (voltage) at the point P.

ChargedRod

The above Figure shows the rod. Highlighted is one small piece ofthe rod which is located at a position x and has length (dx). Note:In answers below, if you want to include dx, it is good practice toput (dx) in parentheses.

(a) Suppose that the rod is uniformly charged to a value Q. Writethe expression for the linear charge density, using the symbolsalready defined.
lambda = 1Click here to preview your answer.

(b) The small piece of the rod has length dx. In terms of symbolsL, x, (dx), D, Q and constants, what is the infinitesimal charge onthe rod? You may need to enclose dx in parentheses for it todisplay properly.
dq = 2Click here to preview your answer.

c) What is the distance from the point to the small piece of therod? Answer in terms of symbols x, L, D, (dx) and constants.
3Click here to preview your answer.

(d) What is the infinitesimal voltage produced at point P from thisdq? Answer in terms of symbols k, Q, x, L, D, (dx) andconstants.
dV = 4Click here to preview your answer.

(e) The final step would be to integrate this result. What are thelower 5Click here to preview your answer.
and upper 6Click here to preview your answer. limits on theintegration? Answer in terms of symbols x, L, D, (dx) andconstants.

Note that if the charge of the rod had not been uniform, we wouldneed to know lambda as a function of x and use that in place of(Q/L).

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