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Electric Potential due to Uniformly Charged Rod At Equatorial Point

Electric Potential due to Uniformly Charged Rod At Equatorial Point :- Consider a uniformly charged thin rod of total length and total charge . The linear charge density is :

$\displaystyle \lambda =\frac{Q}{L}$

Let the rod be centered at the origin and placed along the x-axis, extending from to . Let us determine the electric potential at a point located at a perpendicular distance above the midpoint of the rod (on the y-axis).

Expression for Electric Potential due to Uniformly Charged Rod At Equatorial Point

The electric potential at a point P due to a small :

$\displaystyle dV=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{dq}{\sqrt{{{r}^{2}}+{{x}^{2}}}}$

Where is the charge on a small element of length

To find the total potential, we have to integrate over the entire length of the rod from x = – L/2 to x = + L/2 :-

$\displaystyle V=\int{dV=}\frac{1}{4\pi {{\varepsilon }_{0}}}\int_{{}^{-L}\!\!\diagup\!\!{}_{2}\;}^{{}^{+L}\!\!\diagup\!\!{}_{2}\;}{\frac{dq}{\sqrt{{{r}^{2}}+{{x}^{2}}}}}=\frac{\lambda }{4\pi {{\varepsilon }_{0}}}\int_{{}^{-L}\!\!\diagup\!\!{}_{2}\;}^{{}^{+L}\!\!\diagup\!\!{}_{2}\;}{\frac{dx}{\sqrt{{{r}^{2}}+{{x}^{2}}}}}$ …..(1)

Put $\displaystyle x=r\tan \theta \Rightarrow dx=r{{\sec }^{2}}\theta d\theta$ in equation (1) we get :-

$\displaystyle V=\frac{\lambda }{4\pi {{\varepsilon }_{0}}}\int\limits_{{}^{-L}\!\!\diagup\!\!{}_{2}\;}^{{}^{+L}\!\!\diagup\!\!{}_{2}\;}{\frac{r{{\sec }^{2}}\theta d\theta }{{{\left( {{r}^{2}}+{{r}^{2}}ta{{n}^{2}}\theta \right)}^{1/2}}}}$

$\displaystyle \Rightarrow V=\frac{\lambda }{4\pi {{\varepsilon }_{0}}}\int\limits_{{}^{-L}\!\!\diagup\!\!{}_{2}\;}^{{}^{+L}\!\!\diagup\!\!{}_{2}\;}{\frac{r{{\sec }^{2}}\theta d\theta }{r{{\left( 1+ta{{n}^{2}}\theta \right)}^{1/2}}}}=\frac{\lambda }{4\pi {{\varepsilon }_{0}}}\int\limits_{{}^{-L}\!\!\diagup\!\!{}_{2}\;}^{{}^{+L}\!\!\diagup\!\!{}_{2}\;}{\frac{{{\sec }^{2}}\theta d\theta }{{{\left( {{\sec }^{2}}\theta \right)}^{1/2}}}}$

$\displaystyle \Rightarrow V=\frac{\lambda }{4\pi {{\varepsilon }_{0}}}\int\limits_{{}^{-L}\!\!\diagup\!\!{}_{2}\;}^{{}^{+L}\!\!\diagup\!\!{}_{2}\;}{\sec \theta d\theta }$

Now as $\displaystyle \int{\sec \theta d\theta =lo{{g}_{e}}(\sec \theta +tan\theta )}+C$ ,

$\displaystyle \therefore V=\frac{\lambda }{4\pi {{\varepsilon }_{0}}}\left[ lo{{g}_{e}}(\sec \theta +tan\theta ) \right]_{{}^{-L}\!\!\diagup\!\!{}_{2}\;}^{{}^{+L}\!\!\diagup\!\!{}_{2}\;}$ …..(2)

Now let us replace θ into x again,

As $\displaystyle x=r\tan \theta \Rightarrow \tan \theta =\frac{x}{r}$

$\displaystyle \Rightarrow \sec \theta =\sqrt{1+ta{{n}^{2}}\theta }=\sqrt{1+\frac{{{x}^{2}}}{{{r}^{2}}}}=\frac{\sqrt{{{r}^{2}}+{{x}^{2}}}}{r}$

Using the values of secθ and tanθ in equation (2) we get,

$\displaystyle V=\frac{\lambda }{4\pi {{\varepsilon }_{0}}}\left[ lo{{g}_{e}}(\frac{\sqrt{{{r}^{2}}+{{x}^{2}}}}{r}+\frac{x}{r}) \right]_{{}^{-L}\!\!\diagup\!\!{}_{2}\;}^{{}^{+L}\!\!\diagup\!\!{}_{2}\;}$

$\displaystyle \therefore V=\frac{\lambda }{4\pi {{\varepsilon }_{0}}}lo{{g}_{e}}\left[ \frac{\sqrt{{{L}^{2}}+4{{r}^{2}}}+L}{\sqrt{{{L}^{2}}+4{{r}^{2}}}-L} \right]$ …..(3)

Electric Potential due to Uniformly Charged Rod At Equatorial Point

Expression for Electric Potential due to Uniformly Charged Rod At Equatorial Point

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