Electric Potential: Solving for Distance d from Origin

AI Thread Summary
The discussion focuses on calculating the electric potential V at a distance d from the origin for a rod with a nonuniform charge density. The derived formula V = k&[L + dln(1 + d/L)] differs from the book's answer, V = k&[L - dln(1 + L/d)]. The discrepancy arises from the logarithmic properties, specifically the negative sign in front of the logarithm in the book's solution. The user recognizes a potential error in their sign manipulation during integration. The conversation emphasizes the importance of careful handling of logarithmic identities in the context of electric potential calculations.
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A rod of length L lies along the x-axis with its left end at the origin and has a nonuniform charge density @ = &x. What is the electic potential a distance d from the origin?

V = k int dQ/x where:

dQ = &x dx so:

V = k& int (x/x+d)dx from (0 to L)
I let u = x+d and x = u-d
so:
V = k& int (1 - d/u)du = (x+d) - dln(x+d) (0 to L)
and I get: V = k&[L + dln(1 + d/L)]

in the book the answer is: k&[L - dln(1 + L/d)]

what am I doing wrong?
 
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look very carefully:

+ dln(1 + d/L) [your solution] = - dln(1 + L/d) [book's solution]

by the properties of logarithms.
 
I know the properties of logarithms, but look:

(x+d) - dln(x+d) from (0 to L)

[(L+d) - dln(L+d)] - [d - dln(d)] = [L - dln(L+d) + dln(d)] = L + dln(d/L+d): so somewhere I must be making a mistake with the signs, but where?
 
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