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archaic
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- Homework Statement
- A rod of length ##L## lies on the x-axis such that its left tip is at the origin. It's charged and has a nonuniform charge density ##\lambda=\alpha x##, where ##\alpha>0##.
What is the potential at a point ##B## on its bisector? (Use the following as necessary: α, k, L, b, and d.)
- Relevant Equations
- $$dV=k\frac{dq}{r}$$
I'm not sure I understand why I need to use ##d##.. Maybe they want me to have the potential be zero at ##A##?
In any case, I have found$$V(B)=\alpha k\int_0^L\frac{x}{\sqrt{b^2+\left(x-\frac{L}{2}\right)^2}}dx+C=\frac{\alpha kL}{2}\ln\left(\frac{\sqrt{L^2+4b^2}+L}{\sqrt{L^2+4b^2}-L}\right)+C$$and$$V(A)=\alpha k\int_0^L\frac{x}{d+x}dx+C=\alpha k\left(L+d\ln\left(\frac{d}{L+d}\right)\right)+C$$
In any case, I have found$$V(B)=\alpha k\int_0^L\frac{x}{\sqrt{b^2+\left(x-\frac{L}{2}\right)^2}}dx+C=\frac{\alpha kL}{2}\ln\left(\frac{\sqrt{L^2+4b^2}+L}{\sqrt{L^2+4b^2}-L}\right)+C$$and$$V(A)=\alpha k\int_0^L\frac{x}{d+x}dx+C=\alpha k\left(L+d\ln\left(\frac{d}{L+d}\right)\right)+C$$
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