Electric field above the centre of a rectangle

[haruspex]

The z in the hyperphysics diagram is the distance from the charge to the point P. I'm guessing the z in post #27 should be y for when we integrate dy

No need to guess. Just study the two integrals in post #27. What do you have to replace the z² inside the second integral (the one from the link) with in order to make it look like the integrand that I quoted just above it?

 

[says]

No need to guess. Just study the two integrals in post #27. What do you have to replace the z² inside the second integral (the one from the link) with in order to make it look like the integrand that I quoted just above it?

the z² inside the second integral would be y²

 

[haruspex]

the z² inside the second integral would be y²

No, that would make the integrand $(x^2+y^2)^{-\frac 32}$. You need $(x^2+y^2+r^2)^{-\frac 32}$

 

[says]

k = 8.98 * 109
λ = 4*10^-6
a = 0
b=0.10m
z = y² + r²

Ez = 2kλz ∫ dx / (z² + x²)^3/2
Ez = 2kλ/y² + r² [ x / (y² + r² + x²)^1/2] (evaluated from 0 to 10)

 

[haruspex]

z = y²+ r²

Closer.

 

[says]

z = √[y² + r²]

 

[says]

k = 8.98 * 109
λ = 4*10^-6
a = 0
b=0.10m
z = √r²-x²-y²

Ez = 2kλ/ √(r²-x²-y²) ∫ dx / (√(r²-x²-y²) + x²)^3/2

 

[haruspex]

z = √[y² + r²]

Yes. What do you get now for the second stage integral (dy)

z = √r2-x2-y2

No.

 

[says]

Ez = 2kλ / √(y² + r²) [ x / [(√(y²+r²)+x²]^1/2] (evaluate x at 0 and 0.10m)

Ez = ∫ 2kλ / √(y² + r²) [ 0.10 / [(√(y²+r²)+0.10²]^1/2] (evaluate at y = 0 and 0.05)

I haven't simplified the second integral yet because I didn't want to make any simple mistakes...

 

[haruspex]

Ez = 2kλ / √(y² + r²) [ x / [(√(y²+r²)+x²]^1/2] (evaluate x at 0 and 0.10m)

Ez = ∫ 2kλ / √(y² + r²) [ 0.10 / [(√(y²+r²)+0.10²]^1/2] (evaluate at y = 0 and 0.05)

I haven't simplified the second integral yet because I didn't want to make any simple mistakes...

You have one too many square roots in each expression.

 

[says]

Sorry, I'm having a little trouble with formatting at the minute

Ez = 2kλ / √(y² + r²) [ x / [(y²+r²+x²)^1/2] (evaluate x at 0 and 0.10m)

Ez = ∫ 2kλ / √(y2 + r2) [ 0.10 / [(y²+r²)+0.10²)^1/2] dy (evaluate at y = 0 and 0.05)

 

[haruspex]

Sorry, I'm having a little trouble with formatting at the minute

Ez = 2kλ / √(y² + r²) [ x / [(y²+r²+x²)^1/2] (evaluate x at 0 and 0.10m)

Ez = ∫ 2kλ / √(y2 + r2) [ 0.10 / [(y²+r²)+0.10²)^1/2] dy (evaluate at y = 0 and 0.05)

Yes, that looks right.

 

[says]

Ez = ∫ 2kλ / √(y² + r²) [ 0.10 / [(y²+r²)+0.10²)^1/2] dy (evaluate at y = 0 and 0.05)

substituting in r=0.10

Ez = ∫ 2kλ / √(y² + 0.10²) [ 0.10 / [((y²+0.10²)+0.10²)^1/2]

I'm not sure how to evaluate this integral though

 

[haruspex]

Ez = ∫ 2kλ / √(y² + r²) [ 0.10 / [(y²+r²)+0.10²)^1/2] dy (evaluate at y = 0 and 0.05)

substituting in r=0.10

Ez = ∫ 2kλ / √(y² + 0.10²) [ 0.10 / [((y²+0.10²)+0.10²)^1/2]

I'm not sure how to evaluate this integral though

Sorry, we went a bit wrong.
The equation from the link was of the form $k\lambda z\int ... = k\lambda/z [...]$.
We used it to solve an integral of the form $k\lambda r \int ...$. So we should have multiplied the link equation through by r/z to get
$k\lambda r\int ... = k\lambda r/z^2 [...]$.
See how that changes your integral in post #43.

 

[says]

The integral now:

Ez = ∫ 2kλr / (y² + r²) [ x / (y²+r²+0.10²)^1/2] dy (evaluate at y = 0 and 0.05)

 

[haruspex]

The integral now:

Ez = ∫ 2kλr / (y² + r²) [ x / (y²+r²+0.10²)^1/2] dy (evaluate at y = 0 and 0.05)

The x should be 0.1, but otherwise yes (or should there be another factor of 2?).
This is solvable, but it is a bit involved.
First substitution: y=r tan(θ). What do you get?

Edit: to make the typing easier, maybe write a instead of the 0.10.

 

[says]

I subbed in r=0.10 first

∫ 718.4 / (y²+0.01)(y²+0.02)^1/2

I put it in a wolfram calculator (read:cheated) and got:

71840tan^-1(7.07107y / √(50*y²+1))
substituting in y=0.05
= 102149.75 N/C

 

[says]

dEz = (kλdx / r²) z/r
Ez = 2kλz ∫ dx / (z² + x²)^3/2

where
z = √(y²+r²)

Ez = 2kλ/z[ x / (y²+r²) + x²)^1/2] (bounds of integration are 0 and 0.10m)

Ez = 2kλr / (y² + r²) [ 0.10 / ((y²+r²)+0.10²)^1/2]

k = 8.98*10⁹
λ = 4.0*10^-6
r = 0.10

Ez = 7184 / (y² + 0.10²) [ 0.10 / ((y²+0.10²)+0.10²)^1/2]

Ez = [7184 / (y² + 0.01)] [ 0.10 / ((y²+0.02)^1/2]

Ez = ∫ [7184 / (y² + 0.01)] [ 0.10 / ((y²+0.02)^1/2] dy

Ez = 71840tan^-1 [ 7.07107y / √(50y²+1)

Ez = 102149.75 N/C

 

[haruspex]

I subbed in r=0.10 first

∫ 718.4 / (y²+0.01)(y²+0.02)^1/2

I put it in a wolfram calculator (read:cheated) and got:

71840tan^-1(7.07107y / √(50*y²+1))
substituting in y=0.05
= 102149.75 N/C

Ok. I haven't checked the numerics.
If you are interested in seeing how the integral can be solved I will post it.

 

[says]

Yes please!

 

[haruspex]

Writing $y=r \tan(\theta)$ and c=a/r:
$\int \frac{rady}{(y^2+r^2)(y^2+r^2+a^2)^\frac 12}=\int \frac{c.d\theta}{(c^2+\sec^2(\theta))^\frac 12}$
$=\int \frac {c.d\sin(\theta)}{(1+c^2\cos^2(\theta))^\frac 12}$
writing s for sin(theta) and f² for a²/(a²+r²):
$=\int \frac{f ds}{(1-f^2s^2)^\frac 12}$
Writing s=sin(φ)/f:
$=\int d\phi=[\phi]$
Then it is just a matter of unwrapping all the substitutions.

 

[TSny]

Ez = 102149.75 N/C

As a rough check of your numerical answer you can compare your answer to what you would get if you lumped all the charge of the plate together as a point charge at the origin.

 

[says]

As a rough check of your numerical answer you can compare your answer to what you would get if you lumped all the charge of the plate together as a point charge at the origin.

I'm not sure I follow?

 

[TSny]

I'm not sure I follow?

If you compressed all the charge of the plate into a point charge at the origin, what would be the magnitude of the E field at the point on the z axis at z = 10 cm?

Would you expect this value to be larger or smaller than the answer you got for the plate?

 

[says]

It would be the same wouldn't it?

I'm not sure how to compress all of the charge of the plate into a point source at the origin though.

 

[TSny]

It would be the same wouldn't it?

I'm not sure how to compress all of the charge of the plate into a point source at the origin though.

You can just imagine that you have a point charge at the origin that has a charge equal to the total charge of the plate.

(1) All the charge in the point charge would be a distance of 10 cm from the point on the z axis at z = 10 cm. Compare that to having the charge spread out on the rectangle. Is all the charge on the rectangle at a distance of 10 cm from the point on the z axis or is most of the charge on the rectangle farther away?

(2) Also, for the rectangle, you had to project out the z-component of the field of each small region of the rectangle. For the point charge, all of the charge produces E field entirely in the z-direction.

Considering (1) and (2), would you expect the field of the plate to be smaller or larger than the field of the point charge?

It is very easy to calculate the E field of the imaginary point charge. So, you can see if your expectation is verified.

 

[says]

1) if the charge is spread out then most of it is further away.

2) field of the plate would be smaller than that of the point charge

 

[TSny]

1) if the charge is spread out then most of it is further away.

2) field of the plate would be smaller than that of the point charge

Yes. Good. Did you calculate E for the point charge and compare it what you got for the plate?

 

[says]

Charge density = Q/A
4*10^-6 = Q/(0.10*0.20)
Q = 4*10^-6 / (0.10*0.20)
Q = 8*10^-8
E = kQ/r²
k=8.98*10^-9
r=0.10
E=718.4/0.01
E=71840 N/C

 

[TSny]

Charge density = Q/A
4*10^-6 = Q/(0.10*0.20)
Q = 4*10^-6 / (0.10*0.20)
Q = 8*10^-8
E = kQ/r²
k=8.98*10^-9
r=0.10
E=718.4/0.01
E=71840 N/C

Looks good. So, what does this tell you about your answer for the field of the rectangle?