Taking V = 0 at infinity, integrate the potential due to individual sources charges on the rod (i.e. use V = fk dq/r) to find an expression for the electrostatic potential at point A, a distance d above the midpoint of the rod. (Treat the rod as a line of charge.) Use integral tables, Wolfram Alpha, your calculator, or whatever you need, but you do need to actually get the result of the integration. Evaluate the result for Q = 1 × 10-⁹ C, L = 0.080 m, and for d = 0.010 m. Compare to 10 point charges in a similar geometry, distributed along the length L, with the same total charge as before, total = 1 × 10-⁹ C, so each point charge has q = Qtotal/10 = 1 × 10-¹⁰ C. Use the same values for L = 0.080 m, and for d = 0.010 m. = Check the analytic result for the continuous source by examining the potential far away, at distances d much greater than the length L of the rod. When d » L, we expect that we can no longer discern the particular size and shape of the charge distribution and that instead it will resemble a point charge. Thus, the expression for the electric potential should also resemble the expression for the electric potential of a point charge. Check this prediction at a few test points by evaluating your expression from part (a) using values of d » L. For example, you might calculate the exact and approximate results using d=1.0 m and again using d =3 m, or any distances that are much larger than the length. Use at least two different test points so you can see that the result has the right dependence on the distance - if you triple the distance, as I did in my suggested distances, what should happen to the potential?

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(d) Check the analytic result for the continuous source by examining the potential far away, at distances d much greater than the length L of the rod. When d >> L, we expect that we can no longer discern the particular size and shape of the charge distribution and that instead it will resemble a point charge. Thus, the expression for the electric potential should also resemble the expression for the electric potential of a point charge. Check this prediction at a few test points by evaluating your expression from part (a) using values of d >> L . For example, you might calculate the exact and approximate results using d=1.0 m and again using d =3 m, or any distances that are much larger than the length. Use at least two different test points so you can see that the result has the right dependence on the distance - if you triple the distance, as I did in my suggested distances, what should happen to the potential?

(a) Taking V = 0 at infinity, integrate the potential due to individual sources charges on the rod (i.e.
use V =
fk dq/r) to find an expression for the electrostatic potential at point A, a distance d
above the midpoint of the rod. (Treat the rod as a line of charge.) Use integral tables, Wolfram
Alpha, your calculator, or whatever you need, but you do need to actually get the result of the
integration.
(b) Evaluate the result for Q = 1 × 10-⁹ C, L = 0.080 m, and for d = 0.010 m.
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(c) Compare to 10 point charges in a similar geometry, distributed along the length L, with the same
total charge as before, total = 1 × 10-⁹ C, so each point charge has q = Qtotal/10 = 1 × 10-¹0
C. Use the same values for L = 0.080 m, and for d = 0.010 m.
(d) Check the analytic result for the continuous source by examining the potential far away, at
distances d much greater than the length L of the rod. When d » L, we expect that we can no
longer discern the particular size and shape of the charge distribution and that instead it will
resemble a point charge. Thus, the expression for the electric potential should also resemble
the expression for the electric potential of a point charge. Check this prediction at a few test
points by evaluating your expression from part (a) using values of d » L. For example, you
might calculate the exact and approximate results using d=1.0 m and again using d =3 m, or
any distances that are much larger than the length. Use at least two different test points so you
can see that the result has the right dependence on the distance - if you triple the distance, as
I did in my suggested distances, what should happen to the potential?
FOR E WORK ONLY: Show that your numerical tests are generally true of your results you
will need to use the fact that for x < 1, ln(√1 + x² + x) ≈ ln(1 + x) ≈ x. Show that this is
numerically true for x=0.01, and even better for x=0.001.
Briefly comment on your results. (Note that we do not want to consider the purely mathematical
limit d→ ∞ here, but a situation in which d is large compared to the length L of the rod.)
Should the exact answer (your answer to part a) be larger than the approximation we just made,
smaller than it, or does that vary by distance? Present a physics-language argument, and then
check mathematically (numerically if you wish, algebraically or graphically if you can.)
(e) Should the potential at A be larger, smaller or equal to the potential at a test point B, a
distance d to the right of the right end of the rod? Justify using a physics argument, then find
an expression for the potential at B and check that the math answer agrees with your physics
one. You may check numerically (using the same values for L, d, and Q as before) or analytically
(comparing the general expressions in terms of variables.)

A thin plastic rod of length L has a positive charge Q uniformly distributed along its length.
*A
d
L
d
*B

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