Question

A charge q is uniformly distributed along an insulating straight wire of length $2l$ as shown in Fig. Find an expression for the electric potential at a point located at a distance d from the distribution along its perpendicular bisector :

• A. $V=\frac{1}{4\pi {\epsilon }_0}.\frac{q}{l}in\left(\frac{\sqrt{l^2+d^2}+l}{\sqrt{l^2+d^2}-l}\right)$
• B. $V=\frac{1}{4\pi {\epsilon }_0}.\frac{q}{2l}in\left(\frac{\sqrt{l^2+d^2}-l}{\sqrt{l^2+d^2}+l}\right)$
• C. $V=\frac{1}{4\pi {\epsilon }_0}.\frac{q}{2l}in\left(\frac{\sqrt{l^2+d^2}+l}{\sqrt{l^2+d^2}-l}\right)$
• D. $V=\frac{1}{4\pi {\epsilon }_0}.\frac{q}{4l}in\left(\frac{\sqrt{l^2+d^2}+l}{\sqrt{l^2+d^2}-l}\right)$

Answer 1

The correct option is C

$V=\frac{1}{4\pi {\epsilon }_0}.\frac{q}{2l}in\left(\frac{\sqrt{l^2+d^2}+l}{\sqrt{l^2+d^2}-l}\right)$

PE due to a small element dx of rod at length x from midpoint of rod :

$dPE=\frac{1}{4\pi ϵ}\frac{qdx}{2l\sqrt{d^2+x^2}}$

total PE = ${\int }_{-l}^l\frac{1}{4\pi ϵ}\frac{qdx}{2l\sqrt{d^2+x^2}}$

taking $cos\theta =\frac{d}{\sqrt{d^2+x^2}}\Rightarrow dx=\frac{d}{se{c}^2\theta }d\theta$ And $x=-l\Rightarrow \theta ={\theta }_1=ta{n}^{-1}(-l/d)$ and $x=l\Rightarrow \theta ={\theta }_2=ta{n}^{-1}\left(l/d\right)$

Therefore ,

PE =${\int }_{{\theta }_1}^{{\theta }_2}\frac{1}{4\pi ϵ}\frac{q}{2l}\frac{cos\theta }{d}\frac{d}{co{s}^2\theta }d\theta ={\int }_{{\theta }_1}^{{\theta }_2}\frac{1}{4\pi ϵ}\frac{q}{2l}sec\theta d\theta =\frac{1}{4\pi ϵ}\frac{q}{2l}{\left|ln\right(sec\theta +tan\theta |}_{{\theta }_1}^{{\theta }_2}=\frac{1}{4\pi ϵ}\frac{q}{2l}ln\left(\frac{\sqrt{d^2+x^2}+l}{\sqrt{d^2+x^2}-l}\right)$