Electric flux through open surface

[gracy]

I know in such type of questions we should try to enclose the charge completely and symmetrically by as many bodies requires as that of the given body.
Let's say if Charge Q is kept at the mouth centre of a hemisphere .Here a hemisphere is given so we know if another hemisphere is placed below it will enclose the charge completely by a sphere.So,the flux through one hemisphere is Q/2ε0
But This was a simple case but when there is a cube I have difficult time in covering the charge completely for example when charge Q is placed at the centre of the edge of a cube.Please guide me.

 

[blue_leaf77]

I have difficult time in covering the charge completely

Why is it difficult if your cube is bigger than the charge distribution?

 

[gracy]

Why is it difficult if your cube is bigger than the charge distribution?

I did not understand .Could you please elaborate this?

 

[blue_leaf77]

Actually it was because I did not completely get your point that I asked you in post #3. Ok you have a cube and you place a charged body in the center of the cube, what difficulty are you facing, do you want to calculate the flux through the cube? Am I visualizing the problem correctly?

 

[gracy]

do you want to calculate the flux through the cube?

No.

I have difficult time in covering the charge completely

As I have done in here

.Here a hemisphere is given so we know if another hemisphere is placed below it will enclose the charge completely by a sphere.

 

[blue_leaf77]

Do you want the upper half of the enclosing surface to be a hemisphere and the lower half to be a half cleaved cube?

 

[gracy]

Do you want the upper half of the enclosing surface to be a hemisphere and the lower half to be a half cleaved cube?

No.

we should try to enclose the charge completely and symmetrically by as many bodies requires as that of the given body.

So enclosing surface has to be cube only

 

[blue_leaf77]

If your charge is in a form of a sphere placed at the origin of the coordinate system, and you want to calculate the flux through a half cube placed above it such that its open surface is centered at the origin and slices the charged sphere in half, the flux through it will be half of that of a complete cube, just as the case for spherical enclosing surface. However it may be more complicated if the charge is not centered in the cube and/or it has irregular shape.

 

[gracy]

Well ,

I have difficult time in covering the charge completely for example when charge Q is placed at the centre of the edge of a cube

In this case I am required total four cubes to cover Q completely,I don't understand ,how?

 

[Dale]

Well ,

In this case I am required total four cubes to cover Q completely,I don't understand ,how?

Take an orange. Cut it into four equal slices. Put it on the center of the edge of a box. How many slices did you have to remove to do that?

 

[gracy]

Put it on the center of the edge of a box

You mean the four slices?

 

[Dale]

You mean the four slices?

As many of the four slices as you can to make something that still looks like an orange.

The question is how many slices must you remove to set the orange on the edge.

 

[gracy]

to set the orange on the edge.

How can I?Edge is linear.

 

[Dale]

That is why you have to take out some slices. How many?

If you cannot do this mentally then physically go cut an orange into four slices and see how many you need to remove to place the remaining slices neatly on the edge.

 

[gracy]

I think you are trying to describe how to visualize the intersection of two planes
https://www.physicsforums.com/attachments/fourcubes-jpg.88863/

 

[gracy]

 

[Dale]

Yea, so clearly it took four cubes.

 

[gracy]

Now could you please explain your orange example

 

[Dale]

No. If I knew an easy way to explain it I would have done so rather than suggest you try a physical example. Besides, you understand the geometry now, so what would be the point?

 

[gracy]

OK.This time I took help of intersection of two planes but what if asks charge Q is placed at the corner of a cube?How would I decide how many cubes it would take to cover the charge completely?

 

[Dale]

Come on gracy. What do you think? You have already figured this out for two cases, use the same reasoning approach that you used for those, and apply it here.

 

[leright]

I think what Dale is suggesting is you imagine the fraction of the volume enclosed by one of the cubes. If the charge is located at the corner of a cube the fraction of the volume enclosed by the cube is 1/8. The reciprocal of that is the number of cubes needed to completely enclose the charge.