Question

A rod of length L lies along the x axis with itsleft end at the origin. It has a nonuniform charge densityλ=αx, whereα is apositive constant.

Calculate the electric potential at point B , which lieson the perpendicular bisector of the rod a distance babove the x axis. (Use alphafor α, k_e fork_e, L, b, and d as necessary.)

Answer 1

Consider the potential at a point \(p\) which coordinates \(x\) and \(y\) The contribution \(\mathrm{dV}(\mathrm{X}, \mathrm{Y})\) due to a segment \(d x\) of the charged rod is,

$$ \begin{aligned} d V &=\frac{k_{e} d x^{\prime} d x^{\prime}}{\sqrt{(x-x)^{2}+y^{2}}} \\ V &=k_{e} \alpha \int_{0}^{L} \frac{x^{\prime} d x^{\prime}}{\sqrt{(x-x)^{2}+y^{2}}} \end{aligned} $$

Sub \(x-x=u\)

$$ \begin{aligned} V &=k_{e} \alpha \int_{-x}^{-x+L} \frac{(u+x) d u}{\sqrt{u^{2}+y^{2}}} \\ &=k_{e} \alpha \int_{-x}^{-x+L} \frac{x d u}{\sqrt{u^{2}+y^{2}}}+k_{e} \alpha \int_{-x}^{-x+2} \frac{u d u}{\sqrt{u^{2}+y^{2}}} \\ &=\left(k_{e} \alpha x \ln \left(u+\sqrt{u^{2}+y^{2}}\right)\right)_{-x}^{L-x}+\left(k_{e} \alpha \sqrt{u^{2}+y^{2}}\right)_{-x}^{L-x} \\ &=k_{e} \alpha x \ln \left(\frac{L-x+\sqrt{(L-x)^{2}+y^{2}}}{-x+\sqrt{(-x)^{2}+y^{2}}}\right)+k_{e} \alpha \sqrt{(L-x)^{2}+y^{2}}-k_{e} \alpha \sqrt{(-x)^{2}+y^{2}} \end{aligned} $$

Sub \(x=-d\) and \(y=0\) The coordinate of point \(\mathrm{B}\) is \(x=\frac{L}{2}\) and \(\mathrm{y}=\mathrm{b}\) Higher order terms are cancel

$$ V\left(\frac{L}{2}, b\right)=k_{e} \alpha \frac{L}{2} \ln \left(\frac{\sqrt{\left(\frac{L}{2}\right)^{2}+b^{2}}+\frac{L}{2}}{\sqrt{\left(\frac{L}{2}\right)^{2}+b^{2}}-\frac{L}{2}}\right) $$