# Homework Help: Check gauss's concepts

1. May 2, 2005

### nishant

take a randomly drawn surface,put some charge inside it,this surface should be having some sharp features.if this thing is a metallic surface then it 's surace will be equipotential,due to this charge density on the sharp points will be the maximum,therefore electric field just outside this surace near the sharp points will be the maximum.But now if we take a gaussian sphere with the charge in it such that this gaussian surface coincides with the sharp points then the electric field at all points equidistant from the centre of the gausian sphere will be the same which does not coincide with the above result that the e.f shpuld be maximum near the sharp points.{NOTE:try to take a symmytric shape having symmytric sharp points while drawing the random model}

2. May 2, 2005

3. May 2, 2005

### ZapperZ

Staff Emeritus
You may want to put a little bit of effort into trying to make what you have in your head into a more understandable form when you write it down. It will require some practice.

Now let's do something similar to what I THINK you're asking. Put a point charge Q at an origin. Now put a metallic SQUARE BOX thin surface that has a center conciding with the origin and the charge. The corners of the box should satisfy your "sharp points" requirement.

Now, think on why what you said about the accumulation of charges at the corners might not be correct. The sharp points are further away from Q and see a smaller field. Even with any field enhancement due to the sharp geometry, we can't easily deduce that there has to be an accumulation of charges there. You need to keep in mind that the induced charges on the outside surface of the metallic box is due to the migration of the opposite charges to the inside surface. The E-field on the inside of sharp pointy geometry isn't similar to the outside of sharp-point geometry.

Thus, if you then construct a larger spherical Gaussian sphere beyond the box, there is no reason yet to expect the metallic box would alter anything.

Zz.

4. May 2, 2005

### OlderDan

The gaussian shpere does not have to have a uniform electric field surrounding it. We use a sphere for a spherically symmetric charge distribution to take advantage of the symmetry of the field. When there is no such symmetry, the field varies over the spherical surface. The Gaussian surface integral still gives us the charge enclosed, but how do you compute the integral when you don't have symmetry?

Gaussian surfaces are constructed to exploit the symmetry when there is one. When there is none, it gets much harder if not impossible to calculate the field exactly. Thats why most of your problems involve some obvious symmetry.

5. May 2, 2005

### nishant

ok,now take a cylindrical object of infinite lengthand having a cross section like that of a sunflower,at the edges of the petal will have greater charge density and by gause law the elctric fild just outside this petal can be calculated by applying gauss law for infinite sheet.but now taking a standard cylinder along the lenth of this flower electric field would be different.take the charge to coincide with the centre of the flower

6. May 2, 2005

### nishant

see I don't know how to draw a picture in this box so I will give u a region which is same as the diagram I want to convey
$$a^2y^2=x^2{a^2-x^2}[tex] and thein the second curve replace x by y and vice versa. the circle may be given by [tex] x^2+y^2=a^2 [tex] 7. May 2, 2005 ### nishant see I don't know how to draw a picture in this box so I will give u a region which is same as the diagram I want to convey [tex] a^2y^2=x^2{a^2-x^2}$$ and thein the second curve replace x by y and vice versa.
the circle may be given by $$x^2+y^2=a^2$$

8. May 2, 2005

### ZapperZ

Staff Emeritus
Oy! I give up!

Zz.

9. May 2, 2005

### nishant

hey man don't loose heart,and use the second last poat by me to draw the figure.

10. May 2, 2005

### nishant

the curve is a^2y^2=x^2{a^2-x^2} ,the second one is obtained by replacing y by x and vice versa and the circle is given by x^2+y^2=a^2

11. May 2, 2005

### OlderDan

Metals are conductors. For a static charge distribution on a metal surface, the electric field MUST be perpendicular to the surface on the outside, and zero on the inside. Otherwise the charges would be moving. Since you know that, you can ALWAYS construct a small gaussian surface enclosing a small region of the metal surface using the well know "pillbox" shape. The field will be perpendicular to the end cap on the outside and parallel to the cylindrical walls. Only the end cap contributes to the integral, so the charge inside is directly related to the field perpendicular to the surface. The overall shape of the surface affects the field distribution, but you can always use this pillbox shape near the surface of a metal to relate the local field to the local charge.

A line of charge is a simple charge configuration. By symmetry the electric field is perpendicular to the line, and is the same magnitude at all points the same distance form the line. The shape that exploits this symmetry is a cylinder whose axis is the line of charge.

12. May 2, 2005

### nishant

OLDER DAN,I think we both r saying the same thing,but in this question I have in the first place taken guass law for 'pillbox' which has its crosssection on the first equation's graph's edge points,due to the edges the charge density there is the maximum and hence by gauss law also the electric field near the edges is max,but by taking the line chrage electric field at that point comes out to be the same as that at any other pint equidistant from the charge-here is where the problem lies.

13. May 2, 2005

### nishant

what happened no one is doing it now?

14. May 2, 2005

### vincentchan

Your question has been answered, digest what people say. Guassian Law is always valid no matter how you choose the surface..

.... a simple graph will help clearify your idea....

Last edited: May 2, 2005
15. May 2, 2005

### OlderDan

You are neglecting the fact that the electric field at a gaussian surface does not depend on the surface you have chosen. You need to draw a diagram. Draw a square representing the cross section of an infinite rectangular rod having a charge distributed over the surface. Assume it is a metal and draw the electric field surrounding that square. You know it must be perpendicular to the surfce everywhere, and you know the charge is not uniform. Now draw a concentric circle that contains the square. Look at the electric field at the surface of the circle. It is far from being uniform, and it is not perpendicular to the surface, yet Gauss's law tells us that if we make a cylinder of some radius surrounding the chargeed rod, the surface integral of the normal component of the electric field over the walls of that cylinder is the same regardless of the radius of the cylinder. Doing the integral would be difficult because of the non-uniformity of the field, but we know the answer is a simple multiple of the charge contained within the cylinder. The integral is the same as if the charge were a line of charge at the axis of the cylinder. That does NOT mean that the field is the same as the field from a line of charge. Many different field distributions can lead to the same integral over the surface. The integral only depends on the charge enclosed.

16. May 3, 2005

### nishant

so even if it is a sphere witha charge enclosed by it being situated at the centre it is not necessary for the electric field at places equidistant from the centre to be the same?

17. May 3, 2005

### OlderDan

You are missing the point, and there is no need to talk about complicated charged surfaces to make the point. Suppose you have a point charge. If you surround the point charge with a sphere centered at the charge, the field at the surface of the sphere will be of constant magnitude and directed radially outward. Because of this symmetry, a spherical surface surrounding the charge centered at the charge gives us a way of caluclating the field strength. So we CHOOSE that surface for convenience.

Now take the same sphere, and move it so that the charge is anywhere inside the sphere- not at the center- I mean anywhere. If can be touching the inside of the sphere if you like. Now the electric field is not symmetric over the surface of the sphere, but the integral of the normal component over the surface of the sphere is exactly the same as it was when the charge was in the center of the sphere.

Now get rid of the sphere and replace it with a box containing the charge. The charge can be absolutely anywhere inside the box. The field has no symmetry over the surface of the box, yet the integral of the normal component over the surface of the box is exactly the same as it was when the charge was in the center of the spherical surface. The point is that no matter where the charge is inside of a closed surface, and no matter what the shape of the surface, or the size of the surface, the integral of the normal component of the field over the surface of any shape is exactly the same as it was when the charge was in the center of the spherical surface.

Furthermore, it does not have to be a point charge. The charge distribution can have any shape. If you construct a surface around it, the integral of the normal component of the field over the surface of any shape is exactly the same for all shapes of surfaces and all shapes of enclosed charge distributions. All that matters is how much charge is surrounded by the shape. Nothing else matters.

Take a look at this. Pay special attention to the middle figure. Then imagine changing the shape of the charge to anything you like that will fit inside the surface. The integral is the same in all cases.