Asked by Collin
An electron traveling at 7.00 x 10^6 m/s enters and passes through a parallel plate capacitor. determine the electric field generated by the capacitor. note that the electron just clears the corner of the positive plate.
PLEASE SHOW ALL WORK!!!
PLEASE SHOW ALL WORK!!!
Answers
Answered by
bobpursley
I suspect some vital info is not posted, such as the distance between the plates, the horizonal length of the path of the plates.
I am guessing the total deflection is the distance between the plates.
I am guessing the total deflection is the distance between the plates.
Answered by
Collin
The distance of the parallel plate capacitor is 2.0 cm and the height it 1.5mm.
Answered by
bobpursley
ok, the force on the electron is E*e, and it is deflected .0015m.
Figure the time it took to go through the field .02m at the given velocity. So you know time, and force, and distance deflected.
F=ma a=F/m
distance=1/2 a*t =1/2 Ee/m * t
solve for E
Figure the time it took to go through the field .02m at the given velocity. So you know time, and force, and distance deflected.
F=ma a=F/m
distance=1/2 a*t =1/2 Ee/m * t
solve for E
Answered by
bobpursley
ok, the force on the electron is E*e, and it is deflected .0015m.
Figure the time it took to go through the field .02m at the given velocity. So you know time, and force, and distance deflected.
F=ma a=F/m
distance=1/2 a*t =1/2 Ee/m * t
solve for E
OOPs,
distance=1/2 a t^2=1/2 Ee/m*t^2
solve for E
Figure the time it took to go through the field .02m at the given velocity. So you know time, and force, and distance deflected.
F=ma a=F/m
distance=1/2 a*t =1/2 Ee/m * t
solve for E
OOPs,
distance=1/2 a t^2=1/2 Ee/m*t^2
solve for E
Answered by
Collin
Thanks, that clears it up a little. But would you be able to possibly plug in the numbers so that I could get a clearer image/reasoning of the answer?
That would be greatly appreciated!
That would be greatly appreciated!
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