When finding the electric field from a sheet of charge, , where does the factor of 2 in the denominator come from?
See example 22-6 in your text. Imagine you have an infinite sheet of charge. Then imagine a Gaussian surface with the sheet going through the center of it, and the two ends of the Gaussian surface parallel to the sheet. This surface could be a box, a cylinder, – it doesn’t matter, because Gauss’ Law is true for any surface. The electric field from the infinite sheet is pointing perpendicular to the sheet, by symmetry. The vector of the Gaussian surface is parallel to on the two ends of the surface, and perpendicular to on the sides of the surface.
Related Questions
- When finding the electric field from a sheet of charge, , where does the factor of 2 in the denominator come from?
- When finding the electric field intensity due to an infinite sheet of charge using GUASSS THEOREM...We?
- When defining the electric field, why must the magnitude of the test charge be very small?
