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. Calculate the electric potential at point B , which lies on the perpendicular bisector of the rod a distance b above the x axis. (Use the following as necessary: α, ke, L, b, and d.)

Answer / Explanation:

For the purpose of clarity, it is to be noted that this question is incomplete due to the fact that the diagrammatic illustration accompanying the question have not been proved.

However, we can find the diagram below.

To be able to properly calculate the electric potential at B, we need to first of all calculate the units if alpha ( ∝)

Therefore:

identifying the units of ∝, we have

λ = c / m = α x = c / m² . m

Therefore, so  α has units of c / m².

Now going back to calculate the electric potential at B, according the diagram below that completes the question, we assume that V = 0 at x = infinity, therefore, we treat V as the sum of the point charge potentials from each bit of charge dq.

Therefore,

V = ∫ K dq/r = ∫ lin(l)(o) K λdx/x + d = ∫ lin(l)(o) K ∝xdx/x + d

Now breaking the above down further,

we have,

         V = K∝ ∫ lin(l)(o) (x + d - d) dx / x + d  =  K∝ ∫ lin(l)(o) (1) dx - d.dx / x + d

         =  K∝ [ ( L - 0) - d ( ln [x + d] ║(ln(L)(0)] = K∝ [ L - d (ln [ L + d] - ln [ d] )]

Therefore,

V = K∝ [ L - d in ( L + d / d).

Therefore, electric potential at point B = V = K∝ [ L - d in ( L + d / d).

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answer:

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explanation: