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Electric Potential Due To A Uniformly Charged Disc

Electric Potential Due To A Uniformly Charged Disc :- Consider a thin, uniformly charged disc of radius lying in the yz-plane, centered at the origin. If the total charge on the disc is the surface charge density will be be :

$\displaystyle \sigma =\frac{Q}{\pi {{R}^{2}}}$

We want to find the electric potential at a point located on the x-axis, at a perpendicular distance from the center as shown in figure below :

Expression for Electric Potential Due To A Uniformly Charged Disc

To calculate the potential, we break the disc into infinitesimal rings of radius and width . Each ring acts like a line of charge and contributes an elemental potential at point .

Small amount of charge on a ring of radius and thickness :

$\displaystyle dq=\sigma .(2\pi ydy)$

Electric potential at due to the ring of radius y :

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

Total electric potential at P can be obtained by integrating dV between the limits y = 0 to y = R :

$\displaystyle V=\int{dV}=\frac{\sigma }{2{{\varepsilon }_{0}}}\int\limits_{0}^{R}{\frac{ydy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}}$

Put $\displaystyle {{x}^{2}}+{{y}^{2}}=u$ , so that $\displaystyle du=2ydy$ ,

$\displaystyle \Rightarrow V=\frac{\sigma }{2{{\varepsilon }_{0}}}\int\limits_{{{x}^{2}}}^{{{x}^{2}}+{{R}^{2}}}{\frac{1}{2\sqrt{u}}}du=\frac{\sigma }{2{{\varepsilon }_{0}}}\left[ \sqrt{u} \right]_{{{x}^{2}}}^{{{x}^{2}}+{{R}^{2}}}$