play phryzby with physics

An Electron Pulled by a Proton

solved CoulombsLaw · NewtonsSecondLaw

Q. An electron and a protonqp=−qe are 1 × 10⁻¹⁰ mrep=1×10−10m apart. Taking the charge of an electron as -1.6 × 10⁻¹⁹ Cqe=−1.6×10−19C and its mass as 9.11 × 10⁻³¹ kgme=9.11×10−31kg, what is the electron's acceleration away from the proton, due to the electrical force between themFnet,e=Fe,ep, in m/s²?

−+electron -1.6×10⁻¹⁹ Cproton ?r = 1×10⁻¹⁰ mFa = ?
Inputs
rep
0.00 m
qe
-1.6e-19 C
me
9.11e-31 kg
Outputs
accelerationae?

How it's solved

Solved with CoulombsLaw, NewtonsSecondLaw

Given

  • rep=1×10−10m
  • qe=−1.6×10−19C
  • me=9.11×10−31kg
  • qp=−qe from “a proton”
  • Fnet,e=Fe,ep from “due to the electrical force between them”

Find ae.

Working
  1. a protonqp=−qe
  2. qp=−(−1.6×10−19)
  3. qp=1.6×10−19C
  4. coulombs law(electron, proton)Fe,ep=k⋅qe⋅qprep2
  5. Fe,ep=8.98755179×109⋅(−1.6×10−19)⋅1.6×10−19(1×10−10)2
  6. Fe,ep=−2.30081326×10−8N
  7. due to the electrical force between themFnet,e=Fe,ep
  8. Fnet,e=(−2.30081326×10−8)
  9. Fnet,e=−2.30081326×10−8N
  10. second law(pulled)Fnet,e=me⋅ae
  11. (−2.30081326×10−8)=9.11×10−31⋅ae
  12. ae=−2.526×1022m/s²
  13. ae≈−2.5×1022m/s²to 2 significant figures
Conclusion

The pull between them is tiny by everyday standards, but the electron is so light that it accelerates enormously; the minus sign says the acceleration is towards the proton.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb
1require_relative "../physics"23question :an_electron_pulled_by_a_proton do4  kind :numerical5  describes "An electron and a proton are %{separation} apart. Taking the charge of an electron as %{electron} and " \6            "its mass as %{mass}, %{acceleration}, in m/s²?"7  given separation: "1e-10 m"8  given electron: "-1.6e-19 C"9  given mass: "9.11e-31 kg"10  asks acceleration: "what is the electron's acceleration away from the proton, due to the electrical force between them"11end
solution.rb
1require_relative "../questions/an_electron_pulled_by_a_proton"2require_relative "../science/electric_charges_and_fields/coulombs_law"3require_relative "../science/laws_of_motion/newtons_second_law"45solve :an_electron_pulled_by_a_proton, using: [ CoulombsLaw, NewtonsSecondLaw ] do6  objects :electron, :proton7  electron.goes(:pulled)89  take separation: electron.separation(from: proton)10  take electron: electron.charge11  take mass: electron.mass1213  deduce_from "a proton", proton.charge => -electron.charge14  deduce_from "due to the electrical force between them", electron.net_force => electron.electric_force(by: proton)1516  take acceleration: electron.acceleration17  answer find_value(:acceleration, in: "m/s2", figures: 2)1819  conclusion "The pull between them is tiny by everyday standards, but the electron is so light that it " \20             "accelerates enormously; the minus sign says the acceleration is towards the proton."21end
playground.rb
1require_relative "../solutions/an_electron_pulled_by_a_proton"23play_with :an_electron_pulled_by_a_proton do4  title "An Electron Pulled by a Proton"5end
The laws it drew on
quantitywrittenunitswithinfrom
chargeqC—CoulombsLaw
coulomb constantkN m² C⁻²—CoulombsLaw
separationrm0 … ∞CoulombsLaw
electric forceFeN—CoulombsLaw
volumevolume——Vectors
massmkg0 … ∞Vectors
speedspeed——Vectors
densitydensity——Vectors
number of molesn——Vectors
angular frequencyangularfrequency——Vectors
workwork——Vectors
currentcurrent——Vectors
temperaturetemperature——Vectors
pressurepressure——Vectors
timets0 … 1e+05Vectors
powerpower——Vectors
total path lengths——Vectors
energyenergy——Vectors
gravitational potentialgravitationalpotential——Vectors
coefficient of frictionμ——Vectors
accelerationam/s²—Vectors
velocityvm/s—Vectors
displacementΔxm—Vectors
angular velocityangularvelocity——Vectors
forceFN—Vectors
angular momentumangularmomentum——Vectors
linear momentumplin——Vectors
electric fieldelectricfield——Vectors
average velocityvavm/s—Vectors
magnetic momentmagneticmoment——Vectors
relative velocityvAB——Vectors
impulseimpulse——Vectors
first vectorA—0 … ∞Vectors
second vectorB—0 … ∞Vectors
angle betweenφ—0 … 3.142Vectors
resultantR—0 … ∞Vectors
resultant directionβ—0 … 3.142Vectors
directionα—-3.142 … 3.142Vectors
x componentAx——Vectors
y componentAy——Vectors
positionxm—Velocity
path lengthdm—Velocity
time takenΔts0 … 1e+05Velocity
average speedcavm/s—Velocity
starting positionx0m—Velocity
final positionxm—Velocity
change of velocityΔvm/s—Acceleration
average accelerationaavm/s²—Acceleration
starting velocityv0m/s—Acceleration
final velocityvm/s—Acceleration
gravitygm/s²—UniformAcceleration
net forceFnetN—NewtonsSecondLaw
weightWN—NewtonsSecondLaw
displacement of electron during pulledΔxem—
position of electron at end of pulledx1,em—
position of electron at start of pulledx0,em—
average velocity of electron during pulledvav,em/s—
time taken of pulledΔts0 … 1e+05
average speed of electron during pulledcav,em/s—
path length of electron during pulleddem—
change of velocity of electron during pulledΔvem/s—
velocity of electron at end of pulled from groundv1,em/s—
velocity of electron at start of pulled from groundv0,em/s—
average acceleration of electron during pulledaav,em/s²—
acceleration of electron during pulledaem/s²—
net force on electron during pulledFnet,eN—
mass of electronmekg0 … ∞
electric force on ground by electronFe,geN—
charge of groundqgC—
charge of electronqeC—
separation of ground from electronrgem0 … ∞
electric force on ground by protonFe,gpN—
charge of protonqpC—
separation of ground from protonrgpm0 … ∞
electric force on electron by groundFe,egN—
separation of electron from groundrem0 … ∞
electric force on electron by protonFe,epN—
separation of electron from protonrepm0 … ∞
electric force on proton by groundFe,pgN—
separation of proton from groundrpm0 … ∞
electric force on proton by electronFe,peN—
separation of proton from electronrpem0 … ∞
mass of protonmpkg0 … ∞
weight on groundWgN—
weight on electronWeN—
weight on protonWpN—
x0≡xi(start of motion)
starting position
x≡xi(end of motion)
final position
v0≡vi(start of motion)
starting velocity
v≡vi(end of motion)
final velocity
t≡Δtm
time
Fe,oo=k⋅qo⋅qoroo2
coulombs law
A=Ax2+Ay2
magnitude from components
di(whole)=di(part 1)+di(part 2)
path in parts
Δtw=Δtp+Δtp
time in parts
ai(part 1)=ai(whole)
steady through its parts
mb≠0
has mass?
R2=A2+B2+2⋅A⋅B⋅cos⁡φ
law of cosines
R⋅sin⁡β=B⋅sin⁡φ
law of sines
R⋅cos⁡β=A+B⋅cos⁡φ
along the first
Ax=A⋅cos⁡α
resolved along x
Ay=A⋅sin⁡α
resolved along y
Δxe=x1,e-x0,e
change of position(pulled)
vav,e=ΔxeΔt
displacement a time(pulled)
cav,e=deΔt
path a time(pulled)
Δve=v1,e-v0,e
velocity gained(pulled)
aav,e=ΔveΔt
velocity gained a time(pulled)
v1,e=v0,e+ae⋅Δt
velocity after a time(pulled)
x1,e=x0,e+v0,e⋅Δt+ae⋅Δt22
position after a time(pulled)
v1,e2=v0,e2+2⋅ae⋅(x1,e-x0,e)
velocity after a distance(pulled)
x1,e-x0,e=(v1,e+v0,e)⋅Δt2
mean of the velocities(pulled)
Fnet,e=me⋅ae
second law(pulled)
Fe,ge=k⋅qg⋅qerge2
coulombs law(ground, electron)
Fe,gp=k⋅qg⋅qprgp2
coulombs law(ground, proton)
Fe,eg=k⋅qe⋅qgre2
coulombs law(electron, ground)
Fe,ep=k⋅qe⋅qprep2
coulombs law(electron, proton)
Fe,pg=k⋅qp⋅qgrp2
coulombs law(proton, ground)
Fe,pe=k⋅qp⋅qerpe2
coulombs law(proton, electron)
qp=−qe
a proton
Fnet,e=Fe,ep
due to the electrical force between them
me≠0
has mass?(electron)
mp≠0
has mass?(proton)
Fnet,e=0
free of forces?(electron, pulled)