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Serway 25.35 A rod of length L lies along the x axis with its left end at the origin. It has a nonuniform charge density λ = αx, where α is a positive constant. (a) What are the units of α? (b) Calculate the electric potential at A. Here's a new one for you. Usually we only concern ourselves with uniform charge distributions. But, in this problem, the charge density changes as you move along the x axis. This makes finding the electric potential a little more difficult, but not much. (a) The units of α can be found by algebraically satisfying the units of λ, given the relationship λ = αx. You know the units of x, they're meters. You also know that a linear charge density, like λ, must have units of coulombs per meter. So, plugging in units rather than numbers... x C m m C m2 (b) Let's start with our general formula for finding the electric potential at some point with respect to a charge distribution. (1) V k e dq r Our first task is to parameterize dq in terms of something that has spatial meaning. Well, each little infinitesimal charge, dq, resides within an infinitesimal space along the x axis, which we'll call dx. The amount of charge in each little chunk of x axis, dx, just depends on the charge density at that location, so: dq dx . But, since this problem is a little more advanced, λ changes based upon how far down the x axis you are: dq dx xdx . Okay, so we now have dq in a form we can use for integration. Now, we need to rewrite the 'r' in Eqn. (1) so we can use it in the integration, too. 'r' just stands for the distance from point A to each little chunk of charge. Since A is located a distance d away from the origin on the negative x axis, and since the location of each little chunk dq is variable, we can write r as: rdx Plugging back into Eqn. (1), and establishing limits of integration to account for all the charge, we obtain: dq V ke ke r V ke x f L xi 0 dq k e r xdx dx x f L xdx dx xi 0 To solve the integral, set u d x . Then, xdx u d du . V k e u d du k u du d u du k u 1 e e uf ui d ln u u f Plugging back in for u: V k e d x 0 d ln d x 0 L L L V k e L d ln d L ln d k e L d ln 1 d L V k e L d ln 1 d u i