The electrical field in a particular space is E = (X + 1.2)1 N/C with x in meters. Consider a cylindrical Gaussian surface of radius 16 cm that is coaxial with the x axis. One end of the cylinder is at x = 0. (a) What is the magnitude of the electricflux through the other end of the cylinder at x = 3.2 m? (b) What net charge is enclosed within the cylinder?

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The flux through any surface is given by the integral:

Where E is the electric field and da is a unit normal vector to the surface.

Using the dot product property:

Where is the angle between the vectors A and B

Therefore ...

Solution Summary

This solution contains over 200 words and calculations to determine the magnitude of the electric flux and net change exists in the cylinder. The net charge enclosed within the cylinder is determined.

... We know that the electric field inside a conductor is ... determine E_1, we consider as a Gaussian surface a cylinder ... A with one of its flat surfaces outside the ...

... law the surface integral of electric field is equal ... For spherically symmetric field the surface integral is ... if r < a charge within the Gaussian surface is +q E ...

... radially inward (type 1) or outward (type0)] of the electric field at radial ... Give electric fields in N/C and charges in Coulombs. Solution: Gaussian surface. ...

... 1) Construct a Gaussian cylindrical surface between the rod and the shell to derive the electric field in the inner space as a function of the distance from ...

... Let us consider an arbitrary point P on the Gaussian surface. Electric field intensity magnitudes at P due to charges Q and Q' are given as : _____ E1 ...

... Find the magnitude of the electric field at a point 1.5×10 ... We are asked to find the electric fields inside, on and out ... 1. To the out side Gaussian surface,. ...

... E.ds = Q/ε0 (1) where E = Electric field intensity on the Gaussian surface, Q = Total charge enclosed by the Gaussian surface. ∫E.ds = E x 2ΠrL. ...

... "Electric flux f emerges from a positive charge and ... i) Now to find the field at a distance r ... shells, consider a concentric spherical Gaussian surface of radius ...

... where E = Electric field on the Gaussian surface, Q = Total charge enclosed by the Gaussian surface (the charge ... but opposite charges on their inner surfaces. ...

... Since the magnitude of the electric field is constant everywhere on the spherical Gaussian surface and is normal to the surface at each point, gauss' law gives ...