(a) Explain the Gaussian surface concept (mathematical closed surface) in Gauss's Law.
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Gaussian surface is a closed surface that encloses the charge distribution. The electric flux through the surface is given by 1/ε0 times the charge enclosed by the surface.
Gauss’s law Integral form
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- Implementing Gauss's law on a cylinder. You're going to find the electric field at a point P due to a cylinder made of an insulating material. Again, we'll assume an infinitely long cylinder, although you can consider a section with finite length L, and radius R. Let P be a distance "D" away from the central axis of the cylinder, which has a charge density p. Make sure to keep D and R straight, in terms of which you are using where. The drawing can help. 1. Draw a suitable Gaussian surface on the picture below. P L 2. Add vectors for då of the gaussian surface, (the infinitesimal area elements) on the drawing. 3. Draw vectors for the electric field caused by L on the Gaussian surface. 4. Write down an equation for p. Plug in to a geometrical equation so that it is in terms of given variables. 5. Write down Gauss's law, and use # 4 to substitute on one side. 6. Simplify the other side of Gauss's law based on the geometry of what you did in #2 and #3. 7. Use an equation for the surface…By using Gauss's Law, show that the electric field in the space between the two conductors of a coaxial cable is given by: E(r) τρι 2πει where "-p" is the linear line charge density (Coulombs / meter) defined for the inner conductor.A hollow sphere made from a non-conducting material is shown below in cross-section. The inner radius is R1, and the outer radius is R2. The material is charged uniformly -ρ.(a) Using Gauss’ Law, find an expression in terms R1, R2, and ρ of the magnitude(measured in N/C) and the direction (away from or towards the center) of the electricfield at a distance r from the center of the sphere, for values R1 < r < R2? (Do all thiswork symbolically - don't use the values of part (b) for this part.) (b) The inner radius is R1 = 1.00 cm, and the outer radius is R2 = 4.00 cm. Thematerial is charged uniformly ρ = -1.70 nC/m3. If the electric potential is 0V infinitely faraway, what is the electric potential at the outer surface (r = R2) of the sphere? Please show full work Thank you!
- A Gaussian surface is a closed surface in three-dimensionalspace through which the electric flux is calculated. Given aspherical Gaussian surface that has a radius of 0.5 meters andencloses 30 electrons.(a) Find the value of the electric flux through this surface.(b) From the calculated value of the electric flux, determine thevalue of the electric field at a distance equal to 0.6 meters fromthe center of the surface.Question 1: Gauss' Law: Electric field from surface charge For this problem assume constants are all in SI units that are not shown. An electric field given by E = (3z³+5)k. Calculate the flux through surface S, with side length a = 2, shown below if the surface is at the following heights: (a) z = 0 (b) z = -2 (c) z = 2 E 2.0 m 1.5 m a (d) The electric field in a region is given by E= a/(b + cx)î, where a = 500N m/C, b = 4.0m, and c 1.0. What is the net charge enclosed by the shaded volume shown below? 1.0 m a 425 y