Gauss' Law - Infinite Sheet of Charge

[jegues]

Homework Statement

Using Gauss' law, derive the expression for the electric field intensity vector of an infinite sheet of charge in free space.

Homework Equations

The Attempt at a Solution

See figure attached for their solution and the figure that goes with the problem.

What do they mean by planar symmetry? Is it simply that the whole surface lies in a plane and is symmetric?

Also it's not obvious to me as to how they developped their answer.

Using Gauss's law,

\oint_{S} \vec{E}\cdot\hat{n}dS = \frac{Q_{enclosed}}{\epsilon_{0}}

First we note that the electric field vectors in space due to the sheet are normal to the sheet.

Now if I enclose the sheet in a rectangular cube, the electric field vectors will be pointing parallel to the normal vectors on the top and bottom of the rectangular cube.

Thus,

E\oint_{S}dS = \frac{Q_{enclosed}}{\epsilon_{0}}

The surface area of the rectangular box, \oint_{S}dS should simply be 2(lh) + 2(lw) + 2(wh), where l=length, w=width, h=height.

I don't see this approaching the answer they provide.

Any ideas on where I'm going wrong, or how I'm misinterpreting their solution?

Thanks again!

 

[dacruick]

I might be missing something but where did you get this rectangle from? If you then enclose this infinite sheet in a cube, you have an infinite cube. I don't really see how there would be an electric field from a charge an infinite distance away.

 

[ehild]

The flux crossing one side of the cube is the scalar product of the electric field with the normal of the surface. There are two sides of the cube perpendicular to the electric field, the other four sides are parallel, resulting in zero flux.

ehild

 

[Redbelly98]

Moderator's note: thread moved to Introductory Physics.

 

[jegues]

The flux crossing one side of the cube is the scalar product of the electric field with the normal of the surface. There are two sides of the cube perpendicular to the electric field, the other four sides are parallel, resulting in zero flux.

ehild

I would say there are two sides parallel, and four sides perpenicular to the electric field.

If I have a infinite long sheet of charge, the electric field vectors are going to point straight up and down.

If I were to put this sheet in the xy plane the electric field vectors would point in the positive & negative z directions.

I agree that there will be 4 sides with no flux flowing out of them, but the portions of the cube above and below the sheet should have flux flowing out of them with their electric field vectors parallel to the vector normal to the top/bottom of the cube.

So basically,

E \oint_{S}dS = \frac{Q_{enclosed}}{\epsilon_{0}}

But what is the surface area of one side of the cube? They denote this surface area as,

S_{0}

So the flux flowing out of one side of the cube is,

ES_{0} = \frac{Q_{enclosed}}{\epsilon_{0}}

So the total flux flowing out of the cube is double that, (i.e. it flows out 2 sides, top and bottom)

2ES_{0} = \frac{Q_{enclosed}}{\epsilon_{0}}

The charge enclosed will be,

\rho_{s} \cdot S_{0}

Thus,

E = \frac{\rho_{s}S_{0}}{2S_{0}\epsilon_{0}} = \frac{\rho_{s}}{2\epsilon_{0}}

I'm fairly certain this is the correct way to think about the problem.

 

[ehild]

Correct.

ehild