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Electric Potential Due To A Charged Ring

Electric Potential Due To A Charged Ring :- This article explores how to compute the electric potential due to a uniformly charged ring at two specific locations :

At the center of the ring
At a point on the axis of the ring

Let us consider a ring of radius R having total charge Q uniformly distributed on it. The liner charge density (λ) of the ring is :

$\displaystyle \lambda =\frac{Q}{2\pi R}$

(1). Electric Potential Due To A Charged Ring At The Center Of The Ring

Each element of the ring is equidistant from the center. Let us take a small charge element on the ring. Electric Potential due to dq at the centre :

$\displaystyle dV=\frac{kdq}{R}$

Since all ‘s are at the same distance from the center, the total potential is :

$\displaystyle V=\int{dV=\int{\frac{kdq}{R}=}}\frac{k}{R}\int{dq}=\frac{k}{R}.Q$

$\displaystyle \therefore V=\frac{kQ}{R}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{R}$ …..(1)

(2). Electric Potential Due To A Charged Ring at a point on the axis

Now consider a point on the x-axis, at a distance from the center of the ring. Each element of the ring (carrying a charge dq) lies at a distance $\displaystyle \sqrt{{{R}^{2}}+{{x}^{2}}}$ from point P.

Electric Potential due to element dl at the point P :

$\displaystyle dV=\frac{kdq}{\sqrt{{{R}^{2}}+{{x}^{2}}}}$

Since the distance is the same for all elements, we can factor out the distance and integrate :

$\displaystyle V=\int{dV=\int{\frac{kdq}{\sqrt{{{R}^{2}}+{{x}^{2}}}}=}}\frac{k}{\sqrt{{{R}^{2}}+{{x}^{2}}}}\int{dq}=\frac{k}{\sqrt{{{R}^{2}}+{{x}^{2}}}}\int{\lambda dl}$