Homework explain

Two parallel bowl 1.40 cm apart space equally and also oppositely charged. An electron is released from remainder at the surface ar of the an adverse plate and also simultaneously a proton is exit from rest at the surface of the confident plate. How far from the an unfavorable plate is the suggest at i m sorry the electron and also proton happen each other?Known:Qe = -1.6 x 10-19 CQp = 1.6 x 10-19 Cme = 9.11 x 10-31 kgmp = 1.6 x 10-27 kg

Homework Equations

a = qE/md = 1/2*a*t2

The effort at a Solution

Acceleration of electronae = qeE / meAcceleration of protonap = qpE / mpDistance of electronde = aet2 / 2Distance that protondp = apt2 / 2Since ns don"t understand time, i tried to eliminate it by t2 = 2de/aeand plugging it right into distance formula yet from there, I"m grounding on what come do.

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Hello, and welcome to PF!You are provided the street D between the plates. Have the right to you make use of this information?
I"m no really sure exactly how distance D relates come the problem, yet my guess would be if electron is travelling from distance 0 to 1.4 cm and the electron is travelling from 1.4 centimeter to 0, if looking at it on an x-axis... Climate the street the electron travels is S, if the street the proton travel is 1.4 centimeter - S.Rewriting the equations would certainly be...S = 1/2 * ae * t2 and0.014 - S = 1/2 ap * t2How would I obtain the acceleration without learning the electric field E?
I"m no really sure exactly how distance D relates come the problem, however my guess would certainly be if electron is travel from distance 0 come 1.4 cm and the electron is travelling from 1.4 cm to 0, if looking in ~ it on an x-axis... Then the street the electron travel is S, if the distance the proton take trip is 1.4 cm - S.Rewriting the equations would certainly be...S = 1/2 * ae * t2and0.014 - S = 1/2 ap * t2
Maybe girlfriend won"t require actual values for the accelerations. Just keep working with symbols. What deserve to you learn about the time from your equations above? because that now, usage D because that the distance between the plates quite than use the number .014 m.
Ah i see.I understand that t would certainly be the exact same for both since that"s when they fulfill with each other. Would certainly I resolve for t and also make them equal to each and also solve for S?t2 = 2S / ae and also t2 = 2(D-S) / apAnd fixing for S would be... S = D * ae / ap + aeUnless my my algebra is wrong.

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Did you median S = D * ae / (ap + ae) so the both ap and also ae space in the denominator?Can you compose your result in regards to the ratio of ae to ap?
Yes they are both in the denominator. Using the formula a = qE/m , I know that E have to be the same for both the proton and also electron, so ns can set them equal.I"m not sure if this is what you were asking however I did(ae * me ) / qe = (ap * mp ) / qpThen resolving for ae i gotae = ap * 1756.31
Yes they space both in the denominator.Using the formula a = qE/m , I understand that E must be the same for both the proton and also electron, so i can collection them equal.I"m not certain if this is what you to be asking but I did(ae * me ) / qe = (ap * mp ) / qpThen addressing for ae i gotae = ap * 1756.31
Ah! i see. Ns plugging in whatever I know, the accelerations obtained cancelled causing S = 0.01399 m. This renders sense since electrons move quicker than protons.