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Electric field at the edges of a cube

  1. Sep 23, 2015 #1
    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 cancelled 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 cancelled 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.
     
  2. jcsd
  3. Sep 23, 2015 #2

    BvU

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    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 ?
     
  4. Sep 23, 2015 #3

    mfb

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    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?
     
  5. Sep 23, 2015 #4
    Yes, but the point was that the angle changes too, which seemed to compensate for the increased distance.

    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.

    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!
     
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