Gaussian surface distribution.

A nonconducting infinite plane sheet of charge has a uniform positive charge per unit area o. To find the electric field E using Gauss's Law, you draw a cylindrical Gaussian surface of length L and cross- the picture. sectional area A that goes through the charged sheet, as shown in E Gaussian X surface X x x X X X X X X X X X X X X X X X X X X VX X XXIX X X + + +++ X Gaussian surface E What is the total electric flux through the Gaussian surface? O = EAL O D = 2EA O Q = EA O D = 2REALD Question 2 3 pts Which TWO statements about Gauss's Law of electrostatics are INCORRECT in the following five options? A larger Gaussian surface will not increase the magnitude of the total electric flux in a charge-free region. O The Gaussian surface cannot be an open surface. The total electric flux going through the Gaussian surface is determined by the enclosed charges and the medium permittivity. O The electric field that is normal to the Gaussian surface should not be taken into account. O) If the medium enclosed by the Gaussian surface has a higher permittivity, the total flux of the electric displacement vector will be always smaller.Question 15 1 pts The figure shows a cubical Gaussian Surface; at the center of the cube is a point charge (the circle) with charge 189. What could explain the flux through BOTTOM face of the Gaussian Surface being 2q,/Eg? TOP BOTTOM An unshown positive charge located above the TOP face of the Gaussian Surface. An unshown negative charge located below the BOTTOM face of the Gaussian Surface. None of the other answers can explain this situation. An unshown positive charge located below the BOTTOM face of the Gaussian Surface. O An unshown positive charge located within the Gaussian Surface