- #36
says
- 594
- 12
z = √[y2 + r2]
Yes. What do you get now for the second stage integral (dy)says said:z = √[y2 + r2]
No.says said:z = √r2-x2-y2
You have one too many square roots in each expression.says said: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...
Yes, that looks right.says said: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)
Sorry, we went a bit wrong.says said: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
The x should be 0.1, but otherwise yes (or should there be another factor of 2?).says said:The integral now:
Ez = ∫ 2kλr / (y2 + r2) [ x / (y2+r2+0.102)1/2] dy (evaluate at y = 0 and 0.05)
Ok. I haven't checked the numerics.says said: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
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 said:Ez = 102149.75 N/C
TSny said: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.
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?says said:I'm not sure I follow?
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.says said: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.
Yes. Good. Did you calculate E for the point charge and compare it what you got for the plate?says said: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
Looks good. So, what does this tell you about your answer for the field of the rectangle?says said: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
Yes, there must be a mistake in your calculation for the rectangle. So, review the calculation to see if you can find an error. Could be a minor mistake somewhere.says said:That my answer is unreasonable because it is greater than that of the point charge calculation
I'm not sure, but it looks to me like you might have a mistake in plugging in the limits of y at the very end. You might check this.says said: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
I think this is OK.says said:Ez = 71840tan-1(7.07107y / √(50y2+1))
Did you take care of both the upper and lower limits of y?substituting y=0.05 into the equation
I don't get this answer when using your expression above with the limits for y.Ez = 102149.75 N/C
Aren't the limits -0.05 m and + 0.05 m? Or did you replace the lower limit by 0 and include a factor of 2?says said:The limits of y are 0.05m and 0 though
OK So, you would havesays said:I replaced the lower limit with 0 and included a factor of 2. :)
TSny said:OK So, you would have
Ez = (2) 71840tan-1(7.07107(.05) / √(50(.05)2+1))
I don't get your answer when I evaluate this.