yhese points are true because it produce same results
If there is no charge inside of a Gaussian surface, the electric field must be zero at points of that surface.
Only charge enclosed within a Gaussian surface can produce an electric field at points on that surface.
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True or false physics 2 questions. 1. [ ] Gauss's law states that the net electric flux ΦE through any closed Gaussian surface is equal to the net charge inside the surface divided by 4πε_0. 2. [ ] Gauss's law is useful for calculating electric field when the charge distribution is highly symmetrical. 3. [ ] At electrostatic equilibrium, the electric field is zero everywhere inside a conductor, and any charge can only be distributed on the surface of the...
(a) We have said that Gauss’s law is always true, but only useful for calculating the electric field created by source charge distributions that are spheres, infinite straight cylinders, and infinite flat sheets, and even those cases have additional restrictions. Explain why we are limited to those distributions. Discuss what additional restrictions apply. For example, can we use Gauss’s law to find the field of a sphere whose density depends on distance r from the center? Can we do it...
Quie 1 06 February, 2018 i. Which of the following laws, principles, or definitions that encloses charge? desch the electric flux through a sartace a) Faraday's law b) Gauss'law c) Kirchoff's law d) Fluxon principal e) Electric Field definition 2. Which of the following statements concerning the electric field inside a conductor is is true? a) The electric field inside a conductor is never zero N/C b) The electric field inside a conductor is always zero N/C. c) The electric...
1) How do you determine the direction of the electric field from afield map? a The electric field is perpendicular to the field lines everywhere. b The field lines point in the direction of the electric field. c The field lines point in the opposite direction of the electric field. d You can not determine the direction of the electric field by the field map alone; the sign of the charge is needed 2) An arbitrarily shaped uncharged conductor is...
The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin/ε0, with Qin/ε, where ε is the permittivity of the material. (Technically, ε0 is called the vacuum permittivity.) Suppose that a 75 nC point charge is surrounded by a thin, 32-cm-diameter spherical rubber shell and...
Gauss' Law and Equipotentials A two-di mensional representation of co-centric spheres and coaxial cables looks the same as below (o). Is this 2-D drawing (a) below for a set of spheres (b) or cyl (a What is your initial guess/prediction? You will measure the electrostatic potential for the 2-D drawing and then compare the data to the predicted potential for the sphere and cylinder configurations above. There are two sections to the lab: calculating/predicting the electrie fields in the two...
Problem 5 Compute the total charge inside in a cylinder of length h and radius Rcy, when ρ(R) αR. Use the result to compute the electric field produced by the cylinder at points outside the cylinder (rRcyl). Note that since > Rcyl, the Gaussian surface (with radius r) encloses all the charge in the cylinder. State the direction of the electric field inside and outside the cylinder when a > 0, that is, when the cylinder carries positive charge. Problem...
#8 Gauss's Law and The Shell Theorem Consider a hollow sphere with charge uni- formly distributed on its surface. Suppose the total charge is Q, where Q may be positive or negative Recall that Gauss's law as we have seen it is: Qenclosed ΣΕ A = EO where A = 47tr2 is the total area of the Gaussian surface Suppose the sphere radius is Ro and r > Ro. In terms of Gauss's Law, the reason why the electric field...
The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin / Eo, with Qin /ɛ, where ε is the permittivity of the material. (Technically, so is called the vacuum permittivity.) Suppose that a 75 nC point charge is surrounded by a thin, 32-cm-diameter spherical...
For Gauss' law where is the Electric field evaluated? O All points in space The center of the charge distribution At an arbitrary point in space On the Gaussian surface o It depends on the symmetry An arbitrarily shaped uncharged conductor is added to a field map. Which field lines are affected by the addition of the conductor? No field lines are affected Only field lines that intersect the conductor are affected Most field lines are somewhat affected