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Calculate potential on the axis of a disc of radius R due to a charge Q uniformly distributed on its surface.

Physics - Exemplar Problems

Chapter 2. Potential and Capacitance



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Suppose we take a circular ring of width dr  and radius r, the charge enclosed with in that area dQ  =  2 π r dr ρ
         ρ is the charge density per unit area = Q /(πR^2)

The potential due to this  dQ at a point P  at distance d from the center of the disc on the axis of the disc is:   dV = (1/4πε₀)  dQ / √(d² + r²)

dV=\frac{1}{4 \pi \epsilon_0} \frac{dQ}{\sqrt{d^2+r^2}}\\\\V= \frac{1}{4 \pi \epsilon_0}\int\limits^{R}_{0} {\frac{2\pi r \rho}{\sqrt{d^2+r^2}}} \, dr\\\\ =\frac{\rho}{2 \epsilon_0} [\sqrt{d^2+r^2}]_0^R\\\\=\frac{\rho}{2\epsilon_0}\ \sqrt{d^2+R^2}\\\\V=\frac{Q}{2\pi \epsilon_0 R^2}\sqrt{d^2+R^2}

i suppose this is right.

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