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Introductory Physics Homework Help
How Is Electric Flux Calculated for a Non-Uniform Field in a Cube?
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[QUOTE="joemama69, post: 2689882, member: 174884"] [h2]Homework Statement [/h2] Consider a cube of sides L. Suppose that a non-uniform electric field is present and is given by E(x) = {a(x+L)^2}[B]x[/B] - (ayL)[B]y[/B] where a is a constant. a) determine the elctric flux through the face of the cube that lies in the xy plane at z = 0 (express answer in a & L b) what is the total flux through the cube c) what is the volume charge density within the cube? d) if L=2 and the total charge within the cube is 70.8pC, determine the value of a [h2]Homework Equations[/h2] [h2]The Attempt at a Solution[/h2] a) determine the electric flux throught face on the xy plane (z=0) first off for the electric field the bold x & y I am assuming are the same as the vector directions i & j so E(x) = {a(x+L)^2}[B]i[/B] - (ayL)[B]j[/B] [tex]\oint[/tex] E dA where E is noted above and dA = [B]k[/B]dxdy = [B]k[/B]2dx therefore the flux =0 because the electric field does not have a k component b) the total flux through the cube so for this i believe i only need to find it throught the surfaces of the xz & yz planes and then double it... dA = dydz[B]i[/B] + dxdz[B]j[/B] = 2dx[B]i[/B] + 2dx[B]j[/B] [tex]\oint[/tex] {a(x+L)^2}[B]i[/B] - (ayL)[B]j[/B] dot dydz[B]i[/B] + dxdz[B]j[/B] from 0 to L = [tex]\oint[/tex] 2a(x+L)^2 dx - 2ayLdx from 0 to L = 2aL(x+L)^2 - 2ayL^2 which i must double to inslde the two oposite faces Flux = 4aL[(x+L)^2 - yL] c) find the volume charge density p = dQ/dV so it seems like i have to diferentiate Q interms of V, but there is not V so I am not sure I am on the right path d) when I plug the values in I am still left with the x & y variables. are these suppose to remain as variables or am i missing something = [/QUOTE]
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How Is Electric Flux Calculated for a Non-Uniform Field in a Cube?
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