8
$\begingroup$

Consider two electrons approaching each other at rather fast speeds, maybe even coming close to colliding. Does gravity play any role in this event? If so, how much influence does it have? Do we need gravity in charged particles?

$\endgroup$
21
$\begingroup$

Gravitation is a negligible effect unless you have a large number of particles and a very small percentage of net charge.

For two protons, the electromagnetic force is $10^{36}$ times stronger than their gravitational force. But if you have a small object made up of protons and neutrons and electrons, with a total of about $10^{36}$ atoms and just one more proton than electrons, then the forces are equal. For planetary, stellar, and galactic sizes the electromagnetic force can be ignored compared to gravity, except when you have localized overcharges like around solar flares.

For atoms and molecular effects, or even electrical circuits, gravitation can just be ignored

See relative strengths of the elementary 4 forces at Wikipedia page on fundamental interaction.

$\endgroup$
  • 3
    $\begingroup$ To emphasize this imbalance, the Sun has a slight charge imbalance, because everything else being equal, electrons are easier to boil off as solar wind because of their lighter mass. And this charge imbalance comes down to.... 77 coulombs, over the Sun's $10^{30}\: \mathrm{kg}$ mass $\endgroup$ – Emilio Pisanty May 1 '17 at 18:07
8
$\begingroup$

For two individual electrons, the electric force will be much more powerful at all points.

For example, suppose they are 1 meter apart, the gravitational force on each electron will be (the fact that they are moving does not change the force):

$$F_G = \frac{Gm_e^2}{r^2} = \frac{6.674\times 10^{-11}\times 8.297\times 10^{-61}}{1^2} = 5.538\times 10^{-71} \:\mathrm N$$

The electric force will be:

$$F_E = \frac{kq_e^2}{r^2} = \frac{8.988 \times 10 ^ 9 \times 2.566 \times 10 ^ {-38}}{1^2} = 2.306 \times 10 ^ {-28}\:\mathrm N$$

We can see that the electric force is several magnitudes higher than the gravitational force.

$\endgroup$
  • 3
    $\begingroup$ Nice. The value of $\sqrt{\frac{k}{G}}$ is $1.15\cdot 10^{10}$ kilograms per coulomb. So if each of the two bodies possesses about $10^{10}$ kilograms of mass for each coulomb of net charge, the two forces (gravitational and electrostatic) will have the same magnitude. $\endgroup$ – Jeppe Stig Nielsen May 1 '17 at 19:28
4
$\begingroup$

Although eletromagnetic force tends to be higher in magnitude then gravitational forces, the presence of negative and positive charges neutralizes the effect of this force... Gravitaional force on other hand, is additive and increases propotionaly in magnitude with mass.

$\endgroup$
  • $\begingroup$ There is no positive charge in a system of two electrons. $\endgroup$ – chepner May 2 '17 at 0:43
1
$\begingroup$

For electrons, the gravitational force is much smaller than the electric forces. To get an idea about how small, let's math. $$F_g = \frac{Gm_1m_2}{r^2}\, , \, F_e = \frac{kq_1q_2}{r^2}\\ \frac{F_g}{F_e} = \frac{Gm_1m_2}{kq_1q_2}$$

Notice, the distance separating the electrons does not matter when comparing the forces. All of these constants are pretty easy to find, so the gravitational force is $2.4\times 10^{-43}$ the size of the electric force. Since we know nothing in the universe to within two parts in $10^{43}$, we may safely ignore gravity.

You mentioned that the electrons were on a collision course. Moving charges, especially fast moving charges, create magnetic fields, and those would be important in your calculations.

$\endgroup$
  • $\begingroup$ My question was simple this: Do electrons possess gravity? If so, how do you really, really know? Where is the experimental proof? No big thing. $\endgroup$ – Douglas D. Beatenhead May 12 '17 at 21:56
  • $\begingroup$ @DouglasD.Beatenhead Since everything else that we've measured in the universe (including [massless] light!) validates the theory, the onus would be on you to provide both theoretical and experimental justification for claiming that electrons are fundamentally different from everything else. Best of luck. $\endgroup$ – user121330 May 12 '17 at 22:33
1
$\begingroup$

Technically: yes (any mass attracts any other mass)
How much: Negligible (gravity is ~$10^{36}$ smaller then EM)
Needed: not for calculations (of your electron-electron collision)

But beware that the question of "is it needed" is quite "wrong": We need information to make accurate calculations or don't need them, we need sunlight to grow plants but:

  • electrons have mass (observation)
  • any mass attracts other mass through gravity (observation)

Gravity acts between two electrons. The question of whether they "need" to be attracted cannot be asked here, it's like asking whether you need to fall down if you jump of a building... well you will. Ask the designer of the universe (whoever it may be) why he needed the electrons to "have mass";)

For further informations, read through the other answers, they are good as well.

$\endgroup$

protected by Qmechanic May 1 '17 at 7:32

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.