# Electric field above the centre of a rectangle

1. Jul 27, 2017

### says

1. The problem statement, all variables and given/known data
Exercise in integration: Using the Electric Field E of a straight line
segment of charge, find the electric field at height 10 cm above the centre of a
rectangle of sides 10cm and 20 cm carrying charge density 4 μC/m2.

2. Relevant equations
µ=Q/L
µ = charge density
Q = total charge
L = length of line charge
μ = charge density = 4 μC/m2
E = 1/4πε0 Q/r2 - Electric Field surrounding point charge.

3. The attempt at a solution
Electric Field surrounding a point charge:
E = 1/4πε0 Q/r2

Electric field at the location of a test charge q due to a small chunk of charge in the line dQ
dE = 1/4πε0 dQ/r2

The amount of charge dQ can be restated in terms of charge density
dQ = μ dx

dE = 1/4πε0 μ dx/r2

Change dx to dθ and sweep different angles and substitute dθ/a for dx/r2

dE = 1/4πε0 μ dθ/a2

dEy = dE cosθ

dEy = 1/4πε0 μ / a cosθ dθ

Ey = ∫ 1/4πε0 μ / a cosθ dθ (with θ going from +θ to -θ)

Can I just do the same thing in the other direction and add the two? i.e.

Ex = ∫ 1/4πε0 μ / a cosθ dθ (with θ going from +θ to -θ)

so

Ey + Ex = electric field at height above the rectangle?

The question said to use the 'Electric Field E of a straight line segment of charge' so I thought it just meant to break it up into two line segments. All I would need to do then is calculate the angles from each side of the rectangle.

2. Jul 27, 2017

### rude man

I think you have the right idea but I'm not sure.
You're supposed to start with the dE field of a straight line of either 10 or 20 cm in length and differential width.
Then integrate (as the problem suggests) this line over the other dimension (if 10 cm line, +/- 10 cm; if 20 cm line, +/- 5 cm.)

3. Jul 27, 2017

### haruspex

I read it as breaking the rectangle into thin parallel strips, treating each strip as a straight line of charge. I.e. just a standard dxdy integration, except that it assumes you already have the result for the first stage of integration.

4. Jul 28, 2017

### says

I tried a different approach.

r = √(x2/4 + y2/4+z2
dQ = λds
ds = (x + y)/ 2

VE = λ / 4πε0 ⋅ ds / √r

E = -∇VE

E = -λ / 4πε0 ( ∂ ds/dx x̄ + ∂ ds/dy ŷ + ∂ ds/dz ẑ )

E = -λ / 4πε0 ( [-xy-y2+4z2 / (x2+y2+4z2) ] x̄ - [x2-xy+4z2 / (x2+y2+4z2) ] ŷ - 4-(x-y) / (x2+y2+4z2)3/2

substitute into the equation:
λ: 4*10-6 c/m2
x = 0.10 m
y = 0.20 m
z = 0.10 m

E = -7.989 x̄ - 3.994 ŷ - 15.973 ẑ V/m

Not sure if I'm making this problem more complicated than it is though

5. Jul 29, 2017

### rude man

You're not following instructions which are to use the expression for the E field from a line segment as basis for the solution. Why not do it that way?

But approaching it from a potential viewpoint would certainly be possible otherwise. I did not check your math though.

6. Jul 29, 2017

### says

So if I use the line segment equation and then just sum up what I get for the x & y line.

The electric field, E, at the point, P, will be equal to the electric field at the line segment of x + the electric field at the line segment of y?

dEz = (kλdx / r2) (z/r)
Ez= kλz ∫ dx / (z2 + x2)3/2
Ez= kλ / z [x / (z2 + x2)1/2]
Ez= kλ / z [b / (z2 + b2)1/2 + a / (z2 + a2)1/2]

for the x line: a=-0.05, b=0.05
for the y line: a=-0.10, b=+0.10
z=0.10
λ =4 μC/m2
k = 8.98 * 109 N m2 C-2

7. Jul 29, 2017

### vela

Staff Emeritus
Do you know the expression for the electric field for a line of charge of length L a height z above the center?

8. Jul 29, 2017

### haruspex

I do not know what you mean by the x and y lines. The charge density is given in terms of per unit area, so the charged object is a rectangular sheet, not just the outline of a rectangle. (Or have you stated the problem incorrectly?)
X and Y are the coordinates of points in the square.
Consider a strip of width dx from (x, -b, 0) to (x, b, 0). What is the charge on that strip? Using the formula for the field due to a uniform line of charge, what field does that exert at (0, 0, z)? What is the component of that field in the z direction?
To get the field due to the whole rectangle, you need to integrate wrt x.

9. Jul 29, 2017

### says

10. Jul 29, 2017

### haruspex

Ok, but, setting a=b, that formula gives you the field normal to a line of length 2a at distance z from its midpoint.
In the present context, z is different. The z axis (where P lies, at (0, 0, h), say) is now from the middle of the rectangle and normal to it. If we consider a line (strip) of charge parallel to the y axis (as I described in post #8), how far is P from the middle of the strip?

11. Jul 29, 2017

### says

I think I was correct in post #4 then.

r = √(x2/4 + y2/4+z2
dQ = λds
dS = (x + y)/ 2

VE = λ / 4πε0 ⋅ dS / √r

E = -∇VE

E = -λ / 4πε0 ( ∂ dS/dx x̄ + ∂ dS/dy ŷ + ∂ dS/dz ẑ )

E = -λ / 4πε0 ( [-xy-y2+4z2 / (x2+y2+4z2) ] x̄ - [x2-xy+4z2 / (x2+y2+4z2) ] ŷ - 4-(x-y) / (x2+y2+4z2)3/2

substitute into the equation:
λ: 4*10-6 C / m2
x = 0.10 m
y = 0.20 m
z = 0.10 m

E = -7.989 x̄ - 3.994 ŷ - 15.973 ẑ V/m

12. Jul 29, 2017

### haruspex

Why the /4?
I have no idea how you get that. How about dS=dxdy?
I assume the √ is a typo.

You still don't seem to understand that you need to integrate over the area of the rectangle. If the calculation in the above post found anything it was just the field due to a point charge.

13. Jul 29, 2017

### says

I found an example, which I believe is similar to my question in this document: http://mlg.eng.cam.ac.uk/mchutchon/chargedPlanes.pdf

It uses integration and Gauss' Law to find the magnitude of the electric field. I've redefined a few terms below and attempted to complete my problem again.
Ep: Electric field at point, P.
σ = charge density = 4*10-6 C/m2
r = 10cm
x = 10cm
y = 5cm
Ep = ∫ ∫ σr / 4πε0 sinθ dxdy
d = √ (x2 + y2 + r2)
sinθ = r / √ (x2 + y2 + r2)

Ep = σr / 4πε0 ∫ ∫ 1 / √ (x2 + y2 + r2)3/2 dxdy
Ep = σr / 4πε0 ∫ 2 / (y2 + r2)3/2 dy
Ep = σr / 4πε0 * 2π/r
Ep = σ / 2ε0

The strength of the field is not dependant on how high we are above the rectangle. Anywhere above the plate and the field will be the same. The only thing the electric field at a point, P, is dependant on is the charge density, σ.

Ep = 4*10-6 / 2(8.85*10-12)
Ep =2.26*106 N/C

14. Jul 29, 2017

### SammyS

Staff Emeritus
As @rude man said:
You continue to ignore this instruction which appears in the given problem statement.

Last edited: Jul 29, 2017
15. Jul 29, 2017

### rude man

I think it's important that you see this the way you're supposed to. And it makes things clearer and easier to understand.

1. You were to start with just a line and a point above the line. That gave you the E field at the point. Not a hard thing to do except maybe one integration which you can solve using some math program or a table. Or you can find the solution on the Web; this part is supposed to be a "given" anyway.

2. Now you're dividing the plane into many strips like the one you just solved. The point is still above the middle of each strip; the only difference is that, as you go across the plate,
(a) the distance from each strip varies, and
(b) the direction of the E field generated by each strip also varies.

So just integrate the E field due to all the strips, taking into account the above two facts. You do not start with a differential area. You start with a differential strip.

16. Jul 29, 2017

### haruspex

This is a confusion.
In your equations, x and y are coordinates of a general point in the plane. The 10cm and 5cm give their ranges of values (-10cm, +10cm), (-5cm, +5cm). These should become the bounds on your integrals.
The link you found is only for an infinite sheet. Yours is bounded, so the integrand is right but the bounds are different.

However you have made progress. At least now you do have a dxdy integral.

As @rude man notes, you are instructed to perform the first of the two integrations by using the known result for a finite line of charge.

17. Jul 29, 2017

### says

Ok, so using the equation for a finite line of charge.

a: distance from the centre of the line of charge to point, P. This is given in the problem statement as 10cm=0.10m
L: half the distance of each line. So for one line of the rectangle it is 10/2 = 5cm=0.05m and the other it is 20/2 = 10cm=0.10m
λ: charge density = 4*10-6 C/m2
k: 8.98 * 109 N m2 / C-2

dEz = kaλdx / (x2 + a2)3/2
Ez = ∫ 2kaλdx / (x2 + a2)3/2) *bounds of integration will be 0 to L. Multiplying by 2, which is in the equation.
Ez = 2kaλ ∫ dx / (x2 + a2)3/2)
Ez = 2kλL / (a√(L2+a2))

Ez = 2kλL / (a√(L2+a2))

So for the line of charge along the x-axis, L=20/2=10cm=0.10m
∴ Ez = 2 * (8.98 * 109) * (4 * 10-6) * 0.10 / (0.10√(0.102+0.102))
Ez = 5.08*105 N/C x̄
This is the x component of the electric field at the point, P.

For the y component:
Ez = 2 * (8.98 * 109) * (4 * 10-6) * 0.05 / (0.10√(0.052+0.102))
Ez = 3.21*105 N/C ŷ

E field at point P = 5.08*105 x̄ + 3.21*105 ŷ N/C

18. Jul 29, 2017

### haruspex

That given distance is from the centre of the rectangle, so it will only be from the middle of a line of charge if it is a line through the middle of the rectangle. You need to consider any line parallel to one side of the rectangle.
Draw yourself a diagram. Try to answer my question in post #10.

19. Jul 29, 2017

### says

I'm a little confused with what you're trying to say here and in post #10. I've found another reference, which i believe is similar to my problem so I might look at that and then try again. ()

20. Jul 29, 2017

### haruspex

That diagram helps, except that marking the point outside the plane as X is confusing. Let's call it P.
The "strip" I keep referring to is marked with dashed lines in the diagram.
What is the total charge on that strip?
How long is the strip?
How far is the centre of the strip from P?
If you can answer those questions you can use your formula for the field at P resulting from the strip.
However, that field has a component in the y direction. You need to extract the component in the z direction.