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## Homework Statement

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}

**x**- (ayL)

**y**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

## Homework Equations

## The Attempt at a Solution

a) determine the electric flux throught face on the xy plane (z=0)

first off for the electric field the bold x & y im assuming are the same as the vector directions i & j so E(x) = {a(x+L)^2}

**i**- (ayL)

**j**

[tex]\oint[/tex] E dA where E is noted above and dA =

**k**dxdy =

**k**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

**i**+ dxdz

**j**= 2dx

**i**+ 2dx

**j**

[tex]\oint[/tex] {a(x+L)^2}

**i**- (ayL)

**j**dot dydz

**i**+ dxdz

**j**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 im not sure im 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

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