# I have a rectangular sheet of given dimensions say L wide and H high.

1. May 30, 2010

### dhruv.tara

I have a rectangular sheet of given dimensions say L wide and H high. Out of each point some variable field is flowing through given by rules as follows:

Consider the top left corner of the sheet, the field is perpendicular at that point. As one moves from top left corner to the top right corner the field remains same in magnitude but changes in direction such that at the top right corner it is horizontal to the surface. The change in the direction is uniform.
This happens in whole of the sheet, that the field is perpendicular to the sheet at the left most side and gradually becomes horizontal at the right most ride.

We need to find the flux coming out of the sheet. (NOTE: This is not a homework problem, or I would have posted it in the homework section)

My Steps:

1) I first of all considered the equation of such a field. Since I am interested only in the perpendicular component of field, I made the equation of perpendicular component only.

Such a field V maybe given by V(y) = Asin(pi/2 - (x*pi/2L))
where V(y) is the perpendicular component of the field and A is the max of field. See that at x = 0, field is wholly perpendicular and at x=L field is wholly horizontal.

2) Next I considered a small area element on the sheet of H*dx because a vertical strip (may) simplify my calculations (as the field is constant each vertical strip)

3) The flux through such an element is given by
H * Asin(pi/2 - (x*pi/2L)) dx

4) We obtain the net flux by integrating over the region r as

int(H * Asin(pi/2 - (x*pi/2L)) dx) from 0 to L

which gives -2LAH/pi

Since I don't have any answer to the problem please tell me if my solution is correct, otherwise point the mistakes in it. Any other suggestions are welcomed.

2. May 31, 2010

### Staff: Mentor

Re: Flux

Looks good to me, except for the minus sign. (Check your integration.)

3. May 31, 2010

### dhruv.tara

Re: Flux

Hmm yea I think I did a mistake on that minus sign...

Also can you please tell me how will I approach the same question taking a horizontal strip instead?

4. Jun 1, 2010

### Staff: Mentor

Re: Flux

Since the flux would vary along its length, a horizontal strip would not be a good choice.

5. Jun 1, 2010

### dhruv.tara

Re: Flux

yea I know... Actually the problem gets tougher if the flux is varying on both the sides... Now instead of just the variation along length the same variation is found in breath too... Hence the vertical field exists only at the left top most edge and it falls as mentioned both side...

Line1 : Vertical ---> Horizontal. (horizontal at the right most point only)
Line2: Unit less than vertical ---> Horizontal (horizontal just before the right most point)
Line3: 2 units less than ver --> Hor (2 units before the right most point)
and so on...

This makes the field come like in a upper triangular region only, as the field is 0 in the lower triangular of the sheet.

In such a case the field is varying on both the axis...

How do I approach such a problem?

6. Jun 1, 2010

### Bob S

Re: Flux

I am not sure I completely understand the problem.

Is the sheet non-conducting and uncharged?
Is the sheet in the x z plane; i.e.; 0 < x < L, 0 < z < H?
Is the electric field magnitude constant; Ex2 +Ey2 + Ez2 = constant?
Is the y component of the field Ey perpendicular to the sheet Ey(x,z) = Asin(pi/2 - (x*pi/2L)); i.e., independent of z?

Does the field satisfy Div E = 0 and Curl E = 0?

Bob S

7. Jun 1, 2010

### Staff: Mentor

Re: Flux

I'm not sure I understand your description of how the field varies from point to point. Nonetheless, since the flux per unit area varies in both directions, use a small unit of area (dx dy) instead of a strip. You'll need to set up a double integral.

8. Jun 1, 2010

### dhruv.tara

Re: Flux

Yea... I think I got the double integral part... The difficulty of the problem is to set up the equation of field itself... I'll just try that out and see if I can do it... Thanks for your help