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- PHY102 Physics II © Dr.Cem Özdoğan
- 23-5 Gauss’ Law and Coulomb’s Law
- Figure 23-8 shows a positive point charge q, around which a concentric spherical Gaussian surface of radius r is drawn.
- Divide this surface into differential areas dA.
- The area vector dA at any point is perpendicular to the surface and directed outward from the interior.
- From the symmetry of the situation, at any point the electric field, E, is also perpendicular to the surface and directed outward from the interior.
- Thus, since the angle between E and dA is zero, we can rewrite Gauss’s law as
- This is exactly what Coulomb’s law yielded.
- total area of
- sphere surface
- radius of Gaussian surface