EF and Potential of Disc | Lecture Note - Edubirdie

University:
Santa Fe College
Course:
ETS 2527C | Electromechanical Components and Mechanisms
Academic year:

2024
Views:

47

Pages:

2
Author:

Jabin F.

Calculate the EF and potential at a distance 'z' above the uniformly charged disc of radius R carrying a surface charge density σ. dq = σ ds dE = \frac{dq}{4\pi\epsilon_0(z^2 + r^2)} dE_z = \frac{\sigma r dr}{4\pi\epsilon_0(z^2 + r^2)^{3/2}} \cos\theta d\vec{E} = \frac{\sigma z dr}{2\epsilon_0(z^2 + r^2)^{3/2}} \hat{z} \vec{E} = \int d\vec{E} = \int_0^R \frac{\sigma z dr}{2\epsilon_0(z^2 + r^2)^{3/2}} \hat{z} \vec{E} = \frac{\sigma z}{2\epsilon_0} \left[\frac{1}{z} - \frac{1}{\sqrt{R^2 + z^2}}\right] \hat{z} Limiting Case: Infinite disc (R → ∞): E_z = \frac{\sigma}{2\epsilon_0} \hat{z} At center of disc (z = 0): E_z = 0 Potential: dV = -\frac{dq}{4\pi\epsilon_0\sqrt{(z^2 + r^2)}} = \frac{\sigma 2\pi r dr}{4\pi\epsilon_0(z^2 + r^2)} V = \int dV = \int_0^R \frac{\sigma r dr}{2\epsilon_0(z^2 + r^2)^{1/2}} V = \frac{\sigma}{2\epsilon_0} \left[\sqrt{R^2 + z^2} - z\right] At center (z = 0): V = \frac{\sigma R}{2\epsilon_0}

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