Equilibrium and Electrons (I have the answer, need to know how to do it)

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  • #1
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I have the answer to this question... but I need to know mathematically how to solve it.

Q: An electron is released above the Earth's surface. A second electron directly below it exerts just enough of an electric force on the first electron to cancel the gravitational force on it. Find the distance between the two electrons.

A: 5.07 m
 

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  • #2
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I would know how to solve it if I were given numerical values (I've done three equilibrium problems on my homework, all with numbers, and I solved them without any problems).

The formulas:

Fe = (8.99 x 10^9)(q1)(q2)/ d^2

Fg = (6.67 x 10^-11)(m1)(m2)/ d^2
 
  • #3
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NYROCKFAN said:
I would know how to solve it if I were given numerical values (I've done three equilibrium problems on my homework, all with numbers, and I solved them without any problems).

The formulas:

Fe = (8.99 x 10^9)(q1)(q2)/ d^2

Fg = (6.67 x 10^-11)(m1)(m2)/ d^2

the charge for an electron is a fundamental constant look it up in your textbook it is around 1.6 x 10^-19 Coulombs and the mass of the electron is also a fundamental constant which is 9.11 x 10^-31 kg
 
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stunner5000pt said:
the charge for an electron is a fundamental constant look it up in your textbook it is around 1.6 x 10^-19 Coulombs and the mass of the electron is also a fundamental constant which is 9.11 x 10^-31 kg

I did that... but the answer I got was well over 1000 m
 
  • #5
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Here is what I did... maybe you could tell me where I went wrong:

(8.99 x 10^9)(1.6 x 10^-9)^2 / x^2
=
(6.67 x 10^-11)(9.109 x 10^-31)^2 / x^2
 
  • #6
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Whoops... I put 9 instead on 19.

Let me try that again...
 
  • #7
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Nope... still got a number much bigger than 5
 
  • #8
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Am I not supposed to square the charge and mass values?
 
  • #9
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Ok... tried not squaring them... still a huge number
 
  • #10
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Anyone out there have any ideas?
 
  • #11
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Make it a habit to solve all of your problems symbolically. When you have a symbolic solution, then plug in numbers.

individually your equations are correct. BUT, is the d in the gravity equation the same as the d in Columbs equation? What are the 2 masses in the gravity equation?

I would write the equations

[tex] F_c = K \frac {q_e^2} {d^2} [/tex]
and
[tex] F_g = G \frac {m_e M_E} {{(R_e + D +d)}^2}[/tex]

Where Fc = Coulmb force
Fg = Gravitational force
qe= electron charge
me= mass of the electron
ME= Mass of the Earth
Re=Radius of the earth
D is the distance above the earth
d is the distance between the electrons

K and G are the respective force constants.
edit: added parens in the denominator of the gravitional equation.

Now set the forces equal and solve for d.
 
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  • #12
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I have never seen the second equation in my life... & I have no idea how to use it.

I'm only supposed to use the two equations I've been using.

& I assume the distances are the same because I have no idea what they could be.
 
  • #13
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& about the masses... again I have no idea. I copied the question exactly from my book.

All we learned in this unit so far is stuff like three particles, each has a different charge (which you know), you know the distances of two of them, what is the third distance.
 
  • #14
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Anyone else have any ideas?
 
  • #15
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NYROCKFAN said:
I have never seen the second equation in my life... & I have no idea how to use it.

I'm only supposed to use the two equations I've been using.

& I assume the distances are the same because I have no idea what they could be.
Sure you have, that is simply the distance from the center of the earth to your second electron.

Trouble, we do not know D, the distance of the lower electron from the surface of the earth. Assume it to be zero or nearly so if you need a numerical solution. You also may be able to neglect the d (distance between the electrons) in the gravitational equation.
 
  • #16
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You can find the distance at which the gravitational force of 2 electrons is equal to the Coulomb force, but it is when r = [itex] \infty [/itex]

You need to find the balance between the EARTHS gravitational field and the Coulomb field of the electrons.
 
  • #17
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And what about the gravitational force between the two electrons? Shouldn't it be considered?
 
  • #18
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I am going to finish this problem. Just because.
we have:
Major edit! I am redefining D to be the distance to the upper electron instead of the lower, this removes the extra d greatly simplifing the expression.

[tex] F_c = K \frac {q_e^2} {d^2} = F_g = G \frac {m_e M_E} {{(R_e + D )}^2}[/tex]

We need to isolate d,

[tex]d^2 = \frac {K q_e^2 (R_E + D )^2} {G M_E m_e}[/tex]

[tex] d = q_e (R_E+D) \sqrt {\frac {K} {G M_E m_e} } [/tex]


If we neglect the gravitational force between the electrons and assume that we are near the earths surface this simplifies a to.

[tex] d = q_e R_E \sqrt {\frac {K} {G M_E m_e} } [/tex]

Pluging in the numerical values yields d=5.07m
 
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  • #19
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good idea I will try it
 
  • #20
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I ran out of the time this morning let me explain a little better the effects of D.

[tex] d = (R_E +D) \sqrt {\frac {K} {G M_E m_e} }[/tex]


So here we have an expression for the distance between 2 electrons in gravitational equilibrium with the Coulomb forces of the lower electron which is D m above the earths surface balancing the earths gravitational force on the upper electron. It is clear that as D increases the distance between the electrons will increase. Given this expression you can find the separation at any given distance above the earths surface, you will find that you will need more then 3 significant digits for D < 104m. By setting D=0 we find a minimum separation. Which corresponds to the given result.
 
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  • #21
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Why can we ignore the gravitational force between the electrons?
This force is given by:
[tex] F_e = G \frac {m_e^2} {d^2} [/tex]

This would be added to the Earths gravitational force given by:

[tex] F_E = G \frac {M_E m_e} {R_E ^2} [/tex]

Lets compare the magnitude of these terms by looking at

[tex] \frac {F_e} G [/tex] and [tex] \frac {F_E} G [/tex]

This will just be an order of magnitude comparison so

we have
ME ~ 1024kg
me ~ 10-31kg
RE ~ 106m
d ~ 10

[tex] \frac {F_e} G = \frac {m_e^2} {d^2}\sim 10^{-60}[/tex]

While

[tex] \frac {F_E} G = \frac {M_E m_e} {R_E ^2} \sim 10^{-19}[/tex]

Clearly the gravitational effects between the electrons is much smaller then the earths gravitational effect and can be neglected.
 

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