Electric field at the edges of a cube

In summary, according to the author, the field strength along the edges of a uniformly charged cube will be weaker than over the center of a face, but it is not clear why this is the case. The author argues that the field strength will be stronger along the edges by the following argument. First, the sheet is simplified to a square sheet of charge of length 2a. Next, a point is placed a distance .5a directly above the center of the sheet and the field contribution is explored at that point due to an infinitesimal piece of the sheet located a distance a from the center (in the middle of one of the edges). Finally, the sheet is bent down
  • #1
Sturk200
168
17
We are given a uniformly charged (non-conductor) cube. It is required to understand how the field strength along the edges relates to the field strength over the center of a face.

The correct answer is apparently that the field will be weaker along the edges than over the center of a face, but I am having trouble seeing why.

I convinced myself that the field will be stronger along the edges by the following argument.

First simplify the picture and talk about a square sheet of charge of of length 2a. I want to place a point a distance .5a directly above the center of the sheet and look at the field contribution at that point from an infinitesimal piece of the sheet which is located a distance a from the center (i.e. in the middle of one of the edges). The answer depends on two things: the distance of the charge from the point of interest, which in this case is Sqrt(1.25a^2), and the angle made between the vertical direction and the field line, which here is about 63.4 degrees.

Note that due to symmetry we are interested only in the vertical component, so as the angle gets larger and as the distance gets larger, the field gets smaller.

Now consider the alternative case. We bend the sheet down the center at a right angle and form two sides of a box. Now (like before) we explore a point located .5a directly above the edge of the box, at a 45 degree angle from the extension of either of the sides (i.e. measured radially out from the edge). And we want to find the field contribution at this point due to the same infinitesimal element from before; i.e. due to a point a distance a from the right angle bend, in the middle of one of the edges. Now since the sheet is bent we are interested not in the vertical component of field, but rather in the component in the direction parallel to a line drawn radially out from the corner of the bend, all other components being canceled by symmetry. So how does this radial component of field compare to the vertical component from the previous scenario?

It seems like bending the sheet has two effects: it increases the distance from the point of interest to the element of charge whose influence we are considering (which weakens the field); and it decreases the angle between the field line and the expressed component (which strengthens it). The question is, which of these effects is stronger? On doing some trig, I found that the new distance is about 1.4a and the new angle is about 30.4 degrees, which gives the result that the field due to this element in the bent case is about a tenth stronger than the field due to the analogous element in the flat case. If this is true for one element, I reason that it ought to be true for every element on the sheet.

In sum, bending the sheet increases the distance between charge elements and a point of interest some fixed distance away from the sheet, but it also decreases the proportion of the field component that is canceled by symmetry. It seems that when the sheet is bent by 90 degrees, the second effect outweighs the former by a small amount, the conclusion being that the field at the edge of the cube should be stronger than at the center of a face.

Now numerous people have told me that this result is wrong. I want to believe them. But I also want to know where my reasoning is wrong. Any help is very much appreciated.

Thanks.
 
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  • #2
Hi Sturk,

Nice of you to use so many words to describe a picture, but actually, I'd rather have the picture :smile: .

You know that the field from a charge falls off with 1/distance2, right ?

So if you take your casus 1 and leave the observation point in place, but fold the plate over 90 degrees, don't you think that the average distance charge location --- observation point increases (i.e. the field strength decreases) ?

---

Another tack: for your case 1 you take a charge that's pretty far away from the observation point. But for your case 2 you pick the charge element that's closest by. Not really a level playing field !

---

Third approach: over the center of a face, the field lines go out perpendicular to the surface. Same for the adjacent faces. So at the edges they go out at 45 degrees to the surface. Means they get further and further apart if you move away from those edges. A sure sign of decreasing field strength.

---

In the end you can only convince yourself by actually performing the field calculation. How about setting that up ?
 
  • #3
Sturk200 said:
If this is true for one element, I reason that it ought to be true for every element on the sheet.
Why?
Consider a point 0.01 a away from the center/corner. Its distance increases from ##\sqrt{0.5^2 + 0.01^2}## to ##\sqrt{\frac{0.5^2}{2} + (0.5^2\sqrt{2}+0.01)^2}##, while its angle changes from ##\arctan(0.01/0.5)## to ##\arctan \left( \frac{0.01}{\sqrt{2}(0.5 + \frac{0.01}{\sqrt{2}}} \right)##.
Combined, its effect changes from 3.997 to 3.888, it gets weaker.

And what about the whole interior volume?
 
  • #4
BvU said:
You know that the field from a charge falls off with 1/distance2, right ?

So if you take your casus 1 and leave the observation point in place, but fold the plate over 90 degrees, don't you think that the average distance charge location --- observation point increases (i.e. the field strength decreases) ?

Yes, but the point was that the angle changes too, which seemed to compensate for the increased distance.

BvU said:
Another tack: for your case 1 you take a charge that's pretty far away from the observation point. But for your case 2 you pick the charge element that's closest by. Not really a level playing field !

Sorry if I didn't make it clear, but what I meant was that the test point was the same distance from the charged surface in both cases.

mfb said:
Consider a point 0.01 a away from the center/corner. Its distance increases from 0.52+0.012−−−−−−−−−−√\sqrt{0.5^2 + 0.01^2} to 0.522+(0.522√+0.01)2−−−−−−−−−−−−−−−−−−√\sqrt{\frac{0.5^2}{2} + (0.5^2\sqrt{2}+0.01)^2}, while its angle changes from arctan(0.01/0.5)\arctan(0.01/0.5) to arctan(0.012√(0.5+0.012√)\arctan \left( \frac{0.01}{\sqrt{2}(0.5 + \frac{0.01}{\sqrt{2}}} \right).
Combined, its effect changes from 3.997 to 3.888, it gets weaker.

I think this is exactly the point I was missing. My generalization was too hasty. I guess I need to learn to be cautious of what seems obvious. Thanks for this insight!
 

Related to Electric field at the edges of a cube

What is an electric field?

An electric field is a physical quantity that describes the influence of an electric charge on other charges in its vicinity. It is represented by a vector that gives the direction and strength of the force that a charge would experience if placed in that field.

How does the electric field at the edges of a cube differ from the electric field inside the cube?

The electric field at the edges of a cube is different from the electric field inside the cube because at the edges, the electric field lines are not continuous. This is due to the fact that the electric field lines terminate or originate at the edges, causing a discontinuity in the field.

What factors affect the strength of the electric field at the edges of a cube?

The strength of the electric field at the edges of a cube is affected by the distance from the edge, the charge distribution within the cube, and the orientation of the cube in relation to the source of the electric field.

Can the electric field at the edges of a cube be zero?

Yes, the electric field at the edges of a cube can be zero in certain cases. If the cube is placed in a uniform electric field and is oriented in such a way that the field lines are parallel to the edges, then the electric field at the edges will be zero.

How is the direction of the electric field at the edges of a cube determined?

The direction of the electric field at the edges of a cube is determined by the direction of the electric field lines. The field lines always point in the direction that a positive test charge would move when placed in the field. Therefore, the direction of the electric field at the edges of a cube will depend on the orientation and distribution of charges within the cube.

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