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HELP! Parallel plate Capacitors... |
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| Jan30-05, 12:40 AM | #1 |
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HELP! Parallel plate Capacitors...
Here's the question.
The drawing (attached) shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side. The inital speed of the electron is 7.00x 10^6m/s. The capacitor is 2.00cm long, and its plates are separated by 0.150cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude. Here's what I got. Seems simple enough. I use E=q/(A Eo), where (permittivity of space) Eo=8.85x10^-12C^2/N. Ok so I plug in the charge of an electron divided by the area of the rectangle and Eo. So 1.60x10^-19/((2.00cm/100 x 0.150cm/100)(8.85x10^-12)). I get something like 6.03x10^-8 N/C. See, I didn't use the velocity they gave me, that's where I think the problem is, but I don't see where I could use it! Maybe to find acceleration using kinematics, nope not that... Any ideas? Thanks |
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| Jan30-05, 07:03 AM | #2 |
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| Jan30-05, 12:55 PM | #3 |
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That's odd, then what are the length's and width of the cross-section used for? In terms of acceleration, I don't have time, or distance, but the initial velocity, so I can't use kinematics.
So wait, let's see. Only the horizontal distance is important. And so the d travelled in that velocity is 2.00cm. so t= 0.02m/7.00x10^6m/s = 2.857x 10^-9s(wow that's small) Ok, so I have time, I will plug into x=Vt+1/2at^2. solve for a= 2(x-vot)/t^2. a= 2^-6/(2.857x10^-9s) a=700m/s^2. OK sounds good so far. Now I use F=ma to get force. 9.11x10^-31(700m/s^2)=6.377x10^-28 N and Now I use F=qE, solve for E=F/q. 6.377x10^-28N/(1.60x10^-19) E=1.02x10^-46 N/C... Huh, this doesn't make sense. |
| Jan30-05, 12:57 PM | #4 |
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HELP! Parallel plate Capacitors...
What the, now I get E=F/q and E=3.985x10^-9 N/C...AHHHH
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| Jan30-05, 04:04 PM | #5 |
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Any ideas? Please...
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| Jan30-05, 06:37 PM | #6 |
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No one eh..
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| Jan30-05, 06:50 PM | #7 |
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| Jan30-05, 07:10 PM | #8 |
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Ok,
a=2(0.0015m)/(2.857x10^-9)^2 a=3.675x10^14 F=ma =9.11x10^-31kg x 3.675x10^14 =3.345x10^-16 N. E=F/q =3.345x10^-16N/1.60x10^-19C E=2093 N/C or 2.09x10^3 N/C. NOW that looks like a more sensible answer! Thanks. |
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