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.
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!!!
5 answers
The distance of the parallel plate capacitor is 2.0 cm and the height it 1.5mm.
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
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
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!
1 answer