Electric charge is distributed uniformly along a thin rod of length a, with tota

Question

Electric charge is distributed uniformly along a thin rod of length a, with total charge Q. Take the potential to be zero at infinity.

a) Find the potential at point P, a distance x to the right of the rod (see the figure below). (Use the following as necessary: a, Q, x and0.)

(b) Find the potential at point R, a distance y above the right-hand end of the rod. (Use the following as necessary: a, Q, y and 0.)

(c) In part (a) and part (b), what does your result reduce to as x or y becomes much larger than a? (Use any variable or symbol stated above as necessary.)

part (a)

part (b)

Explanation / Answer

Let the right end of the rod be (0,0) and let the rod lie along x-axis from (-a,0) to (0,0).
(a) Find the potential at point P, a distance x to the right of the rod (see Fig. 23.36). (Use k for the Coulomb constant, and a, Q, x, and y as necessary.)
V(x) = (kQ/a)*[Integral of dp/(p+x); between the limits of p= a to p = 0]; So
V(x) = (kQ/a)*ln[(a+x)/x] = (kQ/a)*ln{1 +( a/x)] ----------------------------------------... 1
____________
(b) Find the potential at point R, a distance y above the right-hand end of the rod.
Similarly
V(y) = (kQ/a)*[Integral of dp/[sq rt[p^2+y^2]] between the limits of p= a to p = 0]; So
V(y) = (kQ/a)[ln[mod{p+sq rt(p^2+a^2)}] (for p =a) -n[mod{p+sq rt(p^2+a^2)} for p =0]
V(y) = (kQ/a)*ln[{a +sq rt(a^2+y^2)}/y] = (kQ/a)*ln[(a/y) + sq rt{1 + (a/y)^2] ------------------ 2

(c) In part (a) and part (b), what does your result reduce to as x or y becomes much larger than a?
part (a)

for x>> 0 tending to infinity V(x) = (kQ/a)*ln[1] = 0

part (b)
for y...0 tending to infinity
V(y) = (kQ/a)*ln[0+sq rt{1+0}] = 0

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