Calculating Charge on Parallel Plates

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An electron accelerates between charged parallel plates, starting with a velocity of 1.63 x 10^4 m/s and reaching 4.15 x 10^4 m/s over a distance of 2.10 cm. The discussion focuses on calculating the charge on the plates, with the user initially struggling with the equations and concepts involved. Key equations include E = (4πkQ)/A and Q = CV, where voltage (V) must be determined from the change in kinetic energy. After several calculations and corrections, the user successfully finds the charge on each plate to be approximately 1.27 x 10^-13 C, confirming the correct answer after extensive troubleshooting. The conversation highlights the importance of understanding energy conservation in electric fields and the relationships between charge, capacitance, and voltage.
  • #31
learningphysics said:
The 1.29E-9 is wrong for capacitance...
AAAH. Fixed that and plugged in again.
YES I got 1.27E-13 for the magnitude of the plates. WHICH IS THE CORRECT ANSWER :D. OMG. It only took 4 hours xD.

I just wanted to thank you for being tenacious enough to stick with the help to the end. I'm so slow with new concepts ><. Thanks alot, I really appreciate it.
 
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  • #32
Etopn23 said:
AAAH. Fixed that and plugged in again.
YES I got 1.27E-13 for the magnitude of the plates. WHICH IS THE CORRECT ANSWER :D. OMG. It only took 4 hours xD.

I just wanted to thank you for being tenacious enough to stick with the help to the end. I'm so slow with new concepts ><. Thanks alot, I really appreciate it.

:smile: You actually meant 1.127E-13 right?
 
  • #33
Etopn23 said:
AAAH. Fixed that and plugged in again.
YES I got 1.27E-13 for the magnitude of the plates. WHICH IS THE CORRECT ANSWER :D. OMG. It only took 4 hours xD.

I just wanted to thank you for being tenacious enough to stick with the help to the end. I'm so slow with new concepts ><. Thanks alot, I really appreciate it.

no prob. :smile: everything takes time, experience, practice etc... just keep doing the problems, and it will become easier...
 
  • #34
learningphysics said:
:smile: You actually meant 1.127E-13 right?
Yeah - Typo sorry xD
 
  • #35
learningphysics said:
no prob. :smile: everything takes time, experience, practice etc... just keep doing the problems, and it will become easier...
Oh I will O_O. Having this many problems on something is unacceptable =\.
 

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