Question
continuous object using integrals. Consider a thin, charged rod of length, L. It lies along the x-axis with one end at x = 0 and the other end at x = L. It has a non-uniform linear charge density given by the function
λ(x) = −λ0 + Ax , where λ0 and A are both positive constants.
(a) What are the units of λ0 and A? Explain.
(b) Suppose that λ0 and A are related such that A = 2λ0/L. What is the charge density of the rod at x = L? Is the left (x = 0) end of the rod positively or negatively charged? Is the right (x = L) end positive or negative?
(c) What is the total charge on the rod? You shouldn’t need to do any calculation. If you’ve understood λ(x) then the answer should be “obvious”.
(d) Consider any point, P, lying somewhere on the x-axis to the right of the rod. It might be helpful to define x at point P as x = L + l. Which way must the electric field at P be pointing? In particular, do you need to worry about calculating both components of the electric field, and is it already clear what the signs of the components must be?
(e) Draw a diagram showing the rod, the axes, and an arbitrary “bit” of the rod. Use this diagram to define the symbols you will need to solve this problem. In particular, you need a variable which specifies the location of each “bit” of the rod which you will eventually integrate over, and an expression for the amount of charge on the “bit”. Don’t cut corners on this diagram! A carefully drawn diagram will help you to understand how to solve the problem and will help the reader (marker) to understand your solution.
(f) Write the approximate E-field due to the one “bit” of rod.
(g) Use your expression for the E-field due to the “bit” to build the integral which will give you the E-field at point P. In particular, be careful of the fact that the two halves of the rod have different signs of charge.
(h) Evaluate your integral any way you want (by hand, using a computer application, using a fancy TI calculator, whatever). You should check the units of your final expression to be sure that you have actually calculated an electric field.

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