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

Question

Electric charge is distributed uniformly along a thin rod of length , with total charge . Take the potential to be zero at infinity. Find the potential at the following points: (see image below) http://i.imgur.com/ISlMX.jpg A)Find the potential at the point P, a distance x to the right of the rod. ans: (kQ/a)ln(1+(a/x)) B)Find the potential at the point R, a distance y above the right-hand end of the rod. ans: (kQ/a)ln((a/y)+(sqrt(1+(a/y)^2))) NEED HELP with the following!! C)In part A, what does your result reduce to as x becomes much larger than a? D)In part B, what does your result reduce to as y becomes much larger than a?

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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