25%. oo T-Mobile Wi-Fi 20:00 pdf (3).pdf Done PHY 2049 Exam 1 ame PID Spring 201

Question

25%. oo T-Mobile Wi-Fi 20:00 pdf (3).pdf Done PHY 2049 Exam 1 ame PID Spring 2015 Problem 4 (25 points) (a) (5 points) Using Gauss's Law, derive the general expression for the electric firid due to an infinite sheet of charge with uniform charge density a Your answer should be expressed in terms of and RA fE Ada Gene Notu Gene. OTA 2E A (b) points) Consider the shown in the figure to the right. The first sheet has charge 0 pchm and lies on the ye-plane. The density of -10 HChm and is second sheet has charge density of parallel to the yr-plane at x 0.10 m. The third sheet has L charge density of ay +10 wcim and is parallel to the yo- Es E plane at x 020 m. Determine the electric field at the point F (0,05 m, 0). E, E la LA Fo Go FIU

Explanation / Answer

Part a:

Gauss law states that the flux through a surface is equal to the charge enclosed by that surface divided by 0. So what exactly is being done is that we are finding out the electric field due to infinite sheet of charges. We are taking a small circular area on the infinite sheet. the electric field through that area will be in the form of cylinder coming out on both the sides of the infinite sheet. There will be only flux through the top and bottom surfaces of the cylinder since the electric field is making angle 00. But through the curved surface the electric field will be perpendicular to the surface normal vector. hence Ecos(90) = 0. so curved surface will not contribute to the total flux. Hence E is uniform for bottom and top so it comes out of integral and only integral of the area is left which gives the total area through which flux comes out. So charge enclosed will be charge density multiplied by the small area taken. hence the field is calculated.

Part b:

We are given three plates and there respective charge density sigma. So the electric field found in part a is /0 so the electric field by the 2nd and 3rd plate will be towards left so it is negative (-i) and due to the 1st plate it is towards right. So field will be E1-E2-E3. plugin the values of for each plates and 0 is common. we get the net field at the point. The point is between 1st and 2nd plate.

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