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Electric Potential Due To Uniformly Charged Rod At Axial Point

Electric Potential Due To Uniformly Charged Rod At Axial Point :- Consider a thin, uniformly charged rod of length , with total charge distributed uniformly along its length. The linear charge density is given by :

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

Let the rod lie along the x-axis, with one end at the origin and the other at . We want to find the electric potential at a point , located at a distance from one end of the rod along the axis of the rod as shown in figure below :

Expression for Electric Potential

The electric potential at point due to a small element dx having charge on the rod is :

$\displaystyle dV=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{dq}{x}$

Where is the charge on a small element of length

To find the total potential, integrate over the entire length of the rod :

$\displaystyle V=\int{dV=}\frac{1}{4\pi {{\varepsilon }_{0}}}\int_{d}^{d+L}{\frac{dq}{x}}=\frac{\lambda }{4\pi {{\varepsilon }_{0}}}\int_{d}^{d+L}{\frac{dx}{x}}$

$\displaystyle V=\frac{\lambda }{4\pi {{\varepsilon }_{0}}}\left[ {{\log }_{e}}x \right]_{d}^{d+L}=\frac{\lambda }{4\pi {{\varepsilon }_{0}}}\left[ {{\log }_{e}}\left( d+L \right)-{{\log }_{e}}\left( d \right) \right]$

$\displaystyle \therefore V=\frac{\lambda }{4\pi {{\varepsilon }_{0}}}lo{{g}_{e}}\left( \frac{d+L}{d} \right)=\frac{Q}{4\pi {{\varepsilon }_{0}}L}lo{{g}_{e}}\left( \frac{d+L}{d} \right)$

Electric Potential due to Uniformly charged rod at Axial Point

Electric Potential Due To Uniformly Charged Rod At Axial Point

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