How Is Electric Flux Calculated for a Non-Uniform Field in a Cube?

[joemama69]

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 I am assuming are the same as the vector directions i & j so E(x) = {a(x+L)^2}i - (ayL)j

$$\oint$$ E dA where E is noted above and dA = kdxdy = k2dx

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 = dydzi + dxdzj = 2dxi + 2dxj
$$\oint$$ {a(x+L)^2}i - (ayL)j dot dydzi + dxdzj from 0 to L

= $$\oint$$ 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
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[anathema]

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Thank you for your interesting question. I am happy to help you with your problem.

a) To determine the electric flux through the face of the cube in the xy plane at z=0, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the total charge enclosed by that surface divided by the permittivity of free space. In this case, the closed surface is the face of the cube in the xy plane at z=0 and the electric flux is given by the electric field times the area of the face. Therefore, we have:

Electric flux = Electric field * Area of face

= {a(x+L)^2}i - (ayL)j * L^2

= aL^2(x+L)^2 - ayL^3

b) The total flux through the cube can be found by adding the flux through each of the six faces. Since the cube is symmetrical, we can just multiply the flux through one face by six. Therefore, the total flux through the cube is:

Total flux = 6 * (aL^2(x+L)^2 - ayL^3)

c) To find the volume charge density, we can use the relation between electric field and charge density, which is given by:

Electric field = Charge density / (2*permittivity of free space)

Therefore, we have:

{a(x+L)^2}i - (ayL)j = p / (2*epsilon)

where p is the volume charge density. Since the electric field is a function of x and y, we can equate the coefficients of i and j on both sides of the equation to get two equations:

a(x+L)^2 = p / (2*epsilon)

-ayL = 0

Solving these equations, we get:

p = 2a(x+L)^2*epsilon

Since the volume charge density is a function of x and y, we cannot determine its value without knowing the values of x and y.

d) To determine the value of a when L=2 and the total charge within the cube is 70.8pC, we can use the relation between electric field and charge density, which we derived in part c). Since the total charge within the cube is given by:

Total charge = Volume charge density * Volume of cube

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