# Flux of a non-uniform E field through the face of a cube

• rudransh verma
In summary: however, you can find the flux through a surface, find the trajectory of a charged particle placed in that field, and so on.
rudransh verma
Gold Member
Homework Statement
The non uniform electric field through a cube is given by E= 3xi^+4j^. We need to find flux of a surface of a Gaussian cube. The surface that we need to find for is at x= 3 meter.
Relevant Equations
Phi= int of(vector E. vector dA)
Phi = int of (3xi^ + 4j^).vector dA = 3 int of(xdA)
Now we put x= 3 and we get at last 36 N m^2/C.
I am getting confused why E is a fuction of x. How can that be? How can we represent the E and position x on the same coordinate system. Is it right ? Because we know distance is inversely proportional to E. how is it multiplied like that?

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rudransh verma said:
Homework Statement:: The non uniform electric field through a cube is given by E= 3xi^+4j^. We need to find flux of a surface of a Gaussian cube. The surface that we need to find for is at x= 3 meter.
Relevant Equations:: Phi= int of(vector E. vector dA)

Phi = int of (3xi^ + 4j^).vector dA = 3 int of(xdA)
Now we put x= 3 and we get at last 36 N m^2/C.
I am getting confused why E is a fuction of x. How can that be? How can we represent the E and position x on the same coordinate system. Is it right ? Because we know distance is inversely proportional to E. how is it multiplied like that?
I'm not100% sure I understand your problem, but I note that you said:
"we know distance is inversely proportional to E"
which doesn't make sense.

Maybe you are thinking about the field a distance d from a point charge q, the magnitude of which is:
##|E| = \frac {kq}{d^2}##
This field is spherically symmetric about the point charge. If this is what you are thinking of, note |E| is inversely proportional to distance squared; conversely, distance is inversely proportional to the square root of |E|.

However, this question says there is a very different sort of field (x-component increasing in proportion to x, constant y-component and no z-component). This sort of field would be produced by a volume charge distribution, i.e. the charge would be spread out in 3D with the charge density being a function of position. This is very different to having a point charge.

The field you're given is the sum of the fields of a distribution of charges. Therefore, the modulus doesn't follow the simple rule for a point charges.

Steve4Physics said:
I'm not100% sure I understand your problem, but I note that you said:
"we know distance is inversely proportional to E"
which doesn't make sense.

Maybe you are thinking about the field a distance d from a point charge q, the magnitude of which is:
##|E| = \frac {kq}{d^2}##
This field is spherically symmetric about the point charge. If this is what you are thinking of, note |E| is inversely proportional to distance squared; conversely, distance is inversely proportional to the square root of |E|.

However, this question says there is a very different sort of field (x-component increasing in proportion to x, constant y-component and no z-component). This sort of field would be produced by a volume charge distribution, i.e. the charge would be spread out in 3D with the charge density being a function of position. This is very different to having a point charge.
I am asking precisely why E and position x are in one coordinate system. How is x multiplied like that. How is this non uniform field? Where is the source located precisely? Can we represent E as a function of x? We know square of x is present in the formula of E in bottom. How did it end up there like that ?

When one writes ##\vec E=3x~\hat i+4~\hat j,## this means that there is a vector field such that the electric field vector has different values at different points in the xy-plane. Anywhere on the y-axis, the y-component is the same. On the x-axis, the x-component increases linearly with distance. Here are some values
##\vec E= [0,4]## at the origin.
##\vec E= [3,4]## at point [1,1].
##\vec E= [-3,4]## at point [-1,-6].
##\vec E= [4.5,4]## at point [1.5,10].
Where the source is located and what shape it has is of no importance. Once you know the field everywhere in space, you have all you need to answer electrostatic questions about what's going on in that region of space. You can find the flux through a surface, you find the trajectory of a charged particle placed in that field, and so on.

Steve4Physics
rudransh verma said:
I am asking precisely why E and position x are in one coordinate system. How is x multiplied like that. How is this non uniform field? Where is the source located precisely? Can we represent E as a function of x? We know square of x is present in the formula of E in bottom. How did it end up there like that ?
It is a non-uniform field because its magnitude changes as x changes. A uniform field would have the same magntiude (and direction) at every point in space.

The source isn't located precisely - it is spread-out, filling all of 3D space, with a charge density that is a function of position.

I can't answer the other questions.

Delta2
rudransh verma said:
We know square of x is present in the formula of E in bottom. How did it end up there like that ?
If the source has finite spatial dimensions then yes it is true that for distance ##r## from the source that is large when in comparison with the spatial dimensions of the source, the electric field is approximately proportional to ##E\approx \frac{C}{r^2}##.

However if the source has some of its three spatial dimensions infinite then the electric field doesn't follow that expression. As an example you can take as source an infinite line of charge, along the x-axis for example that has constant linear charge density ##\lambda##. Most likely this case of infinite source is covered in your textbook, try to find it out.

Anyway this is another interesting problem, that is to find what type of source/charge density generates the electric field ##\vec{E}=3x\hat i+4\hat j##. But for the purpose of the given problem, that is for finding the flux at the face of the given cube, we need not knowing what source generates this field, knowing what expression the field has is enough to answer this question.

rudransh verma and Steve4Physics
Delta2 said:
If the source has finite spatial dimensions then yes it is true that for distance ##r## from the source that is large when in comparison with the spatial dimensions of the source, the electric field is approximately proportional to ##E\approx \frac{C}{r^2}##.

However if the source has some of its three spatial dimensions infinite then the electric field doesn't follow that expression. As an example you can take as source an infinite line of charge, along the x-axis for example that has constant linear charge density ##\lambda##. Most likely this case of infinite source is covered in your textbook, try to find it out.

Anyway this is another interesting problem, that is to find what type of source/charge density generates the electric field ##\vec{E}=3x\hat i+4\hat j##. But for the purpose of the given problem, that is for finding the flux at the face of the given cube, we need not knowing what source generates this field, knowing what expression the field has is enough to answer this question.
And I guess this x has nothing to do with the r^2 term in the expression of E from a point charge.

rudransh verma said:
And I guess this x has nothing to do with the r^2 term in the expression of E from a point charge.
Sure it does. In Cartesian coordinates, the electric field at point ##[x~,y~,z]## due to a point charge ##Q## at the origin is given by $$\vec E=\frac{Q}{4\pi\epsilon_0}\frac{(x~\hat i+y~\hat j+z~\hat k)}{(x^2+y^2+z^2)^{3/2}}.$$ In spherical coordinates, one writes ##\vec r=x~\hat i+y~\hat j+z~\hat k## so that $$\vec E=\frac{Q}{4\pi\epsilon_0}\frac{\vec r}{r^3}=\frac{Q}{4\pi\epsilon_0}\frac{\hat r}{r^2}.$$

Delta2
Steve4Physics said:
It is a non-uniform field because its magnitude changes as x changes. A uniform field would have the same magntiude (and direction) at every point in space.

The source isn't located precisely - it is spread-out, filling all of 3D space, with a charge density that is a function of position.

I can't answer the other questions.
I guess the charge would be the function of position and field would be the function of charge so we can write field in terms of position. Right ? That is what is given here.

rudransh verma said:
I guess the charge would be the function of position and field would be the function of charge so we can write field in terms of position. Right ? That is what is given here.
That's exactly right.

rudransh verma
kuruman said:
That's exactly right.
I have read about such things in motion where a=d^2x/dt^2. That is not what it is. So how is it exactly in unit vector notations.

rudransh verma said:
I have read about such things in motion where a=d^2x/dt^2. That is not what it is. So how is it exactly in unit vector notations.
I don’t know if this helps, but give it a try…

We are given ##\vec E=3x\hat i+4\hat j + 0 \hat k##

We can use this to find the electric field components at any point in space. For example (with coordinates in metres and electric field strength in V/m):

At point ##(-2, 0, 10), \vec E = (-6î + 4î + 0k̂)##
At point ##(-1, 7, 0), \vec E = (-3î + 4î + 0k̂)##
At point ##(0, 2, 19), \vec E = (-0î + 4î + 0k̂)##
At point ##(1, -8, 0), \vec E = (3î + 4î + 0k̂)##
At point ##(2, -3, 98), \vec E = (6î + 4î + 0k̂)##

The field at any point (x, y, z) has:
zero z-component;
a y-component of 4 V/m;
an x-component which is proportional to the point’s x-coordinate (in fact the field's x-components is numerically 3 times the point's x-ccordinate, so in the expression '3x', the '3' has units of V/m² and the 'x' has units of m).

This is an example of a vector field. The magnitude and direction of the field at any point is a function of the point’s coordinates. This field is the result of some (unknown) continuous distribution of charge in space.

kuruman said:
Sure it does. In Cartesian coordinates, the electric field at point ##[x~,y~,z]## due to a point charge ##Q## at the origin is given by $$\vec E=\frac{Q}{4\pi\epsilon_0}\frac{(x~\hat i+y~\hat j+z~\hat k)}{(x^2+y^2+z^2)^{3/2}}.$$ In spherical coordinates, one writes ##\vec r=x~\hat i+y~\hat j+z~\hat k## so that $$\vec E=\frac{Q}{4\pi\epsilon_0}\frac{\vec r}{r^3}=\frac{Q}{4\pi\epsilon_0}\frac{\hat r}{r^2}.$$
Well we don’t know if it’s a point charge. All we know is a field which is a function of position.

kuruman said:
Sure it does. In Cartesian coordinates, the electric field at point ##[x~,y~,z]## due to a point charge ##Q## at the origin is given by $$\vec E=\frac{Q}{4\pi\epsilon_0}\frac{(x~\hat i+y~\hat j+z~\hat k)}{(x^2+y^2+z^2)^{3/2}}.$$ In spherical coordinates, one writes ##\vec r=x~\hat i+y~\hat j+z~\hat k## so that $$\vec E=\frac{Q}{4\pi\epsilon_0}\frac{\vec r}{r^3}=\frac{Q}{4\pi\epsilon_0}\frac{\hat r}{r^2}.$$
You mean if it were a point charge then the eqn of field would be like the first eqn but here it’s probably not a point charge so it’s like that multiplied with 3i^

## 1. What is flux of a non-uniform E field through the face of a cube?

The flux of a non-uniform electric field through the face of a cube is a measure of the amount of electric field passing through a specific area of the cube. It takes into account both the strength and direction of the electric field.

## 2. How is the flux of a non-uniform E field through the face of a cube calculated?

The flux is calculated by taking the dot product of the electric field and the area vector of the cube's face. This accounts for the angle between the electric field and the face of the cube, as well as the magnitude of the electric field at that point.

## 3. Does the shape of the cube affect the flux of a non-uniform E field?

Yes, the shape of the cube can affect the flux. The flux is calculated by taking the dot product of the electric field and the area vector of the cube's face. Therefore, the orientation and size of the face will impact the flux calculation.

## 4. How does a non-uniform E field affect the flux through a cube?

A non-uniform electric field will cause the flux to vary across the face of the cube. This is because the electric field strength and direction are not constant throughout the face, resulting in a non-uniform flux distribution.

## 5. What are some real-world applications of calculating the flux of a non-uniform E field through the face of a cube?

One application is in the design of electronic devices, where understanding the flux of electric fields through different shapes and materials is crucial for proper functioning. Another application is in studying the behavior of electric fields in atmospheric conditions, such as lightning strikes. It is also used in the study of electromagnetic radiation and its effects on objects and materials.

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