- #1
CAF123
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Homework Statement
1) Show that ##\underline{a} = \underline{r} f(r)## is conservative and deduce a functional form for the potential if ##f(r) = r^n##. For what value of n does the potential diverge at both ##\underline{r_o} = 0## and ##\infty##?
The Attempt at a Solution
I have found the potential function but I appear to be out by a constant. f is conservative so take a straight line as the path over integration. So, $$\int_0^{\underline{r}} \underline{a}(\underline{r'}) \cdot d \underline{r'} = \int_0^1 \underline{a}(\lambda \underline{r}) \cdot d(\lambda \underline{r})$$
Inputting ##f(r) = r^n## gives the potential function as ##\frac{1}{2}r^{n+2}##.
When I take the gradient of this, I get ##f(r) \underline{r} (\frac{n}{2} +1)##. This factor of ##(\frac{n}{2} + 1)## is nonexistant in the expression for ##\underline{a}##, so how should I interpret it?
I don't really understand the last part of the question. I would assume that a potential diverging is when it tends to ##\infty##, although this was never defined. When ##r_o = 0, ## I have that $$\int_0^1 r^{n+2} \lambda d \lambda = \frac{1}{2}r^{n+2}$$ while for ##r_o = \infty##, I get$$ r^{n+2} [ \frac{1}{2} - \operatorname{lim}_{t \rightarrow \infty} \frac{t^2}{2}]$$ The second term tends to ∞. I don't think this helps though.
Many thanks.