# Electric fields 2

1. Jul 4, 2009

### leena19

1. The problem statement, all variables and given/known data
A point charge +Q is placed at one of the corners of a cube as shown in the figure.The electric flux through the surface ABCD of the cube due to the charge is
(1) Q (or Q/$$\epsilon_{0}$$)
(2) Q/4 (or Q/4$$\epsilon_{0}$$)
(3) Q/6 (or Q/6$$\epsilon_{0}$$)
(4) Q/24 (or Q/24$$\epsilon_{0}$$)
(5) Q/36 (or Q/36$$\epsilon_{0}$$)

2. Relevant equations

Electric field intensity E=(1/4$$\pi$$$$\epsilon_{0})$$Q/x2

3. The attempt at a solution

I really don't know how to do this.I only know the above equation,but I don't know how to apply it here.
I think a -Q charge would be induced at A,a +Q charge at B,a - Q charge at C and a +Q charge at D,but where do I go from here?

File size:
128.4 KB
Views:
125
2. Jul 4, 2009

### LowlyPion

I can't see the picture, but if the charge is right at the corner, then the total flux of the E-Field will be cut into what ... 8 equal pieces? (Think of stacking 8 cubes with the point charge at the common vertex.)

Now since from Gauss Law you know the total charge of an enclosed surface is Q/εo, then you know that since the cubes are all the same, that 1/8 of the entire field is passing through each cube.

But think about it for a moment and you will see that each cube has 3 external faces (faces opposite the vertex with the charge) and since each face is identical then the flux through any one of them is ...

3. Jul 4, 2009

### leena19

I don't know.
Could you please tell me how you derived this?
I know that,according to Gauss' theorem, the total normal electric flux intersecting a closed surface,of any shape,enclosing a charge Q=Q,
i.e,
A$$\epsilon_{0}E$$ = Q
A is the area enclosed by a charge Q placed at the centre.

I don't undrestand this part,how does the total electric flux divide into 8?

This is the picture,
http://img12.imageshack.us/img12/2296/chargen.png [Broken]

THANK YOU.

Last edited by a moderator: May 4, 2017
4. Jul 4, 2009

### LowlyPion

Imagine 7 other identical cubes arranged all together in a 2 X 2 X 2.

The center vertex will have the charge. (The central vertex is the only spot that all 8 cubes would have in common.) This is where the charge is on your one cube. At a corner.

Since by Gauss Law you know that all the flux through the surface of the 2x2x2 arrangement is Q/εo, then the total of the flux escaping the outward faces of the one cube must be 1/8th of the total.

But they want the flux through just 1 face. A cube has 6 faces. But in the 2x2x2 arrangement you should satisfy yourself that only 3 of the faces are on the outside. Hence the flux through any one face must be 1/3 of what passes out of that cube.

Here's a link for Gauss Law:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html#c1

5. Jul 4, 2009

### leena19

Thank you so much for the detailed explanation.
so the the electric flux through the face ABCD,would be,
[1/(8*3)] Q/$$\epsilon_{0}$$,which would be answer no. (4).

6. Jul 4, 2009

### LowlyPion

Yes. That looks like the answer.

But the understanding is worth much more than just the right answer.

Good Luck.