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Consider two charge particles say $Q_1$ and $Q_2$ both are positive so they creates a electric field around it and both influence each other.

Is it possible that both particles (or charge) creates a Gravitational Field around it, in which intensity varies with distance, if it does then consider another example

Suppose two heavy masses $M_1$ and $M_2$ have very small charges on its surfaces say $dq_1$ and $dq_2$ but comparable and not negligible, so both gravitational and electric fields are generated and one wants to repel and other wants to attract, in this case is the system in equilibrium?

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  • $\begingroup$ There is no question mark in the body of the question! $\endgroup$
    – user249968
    Sep 17 '20 at 8:29
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Can two charged particles exert gravitational force on each other?

Yes. For example, electrons have both mass and charge so they attract each other gravitationally and repel each other electrostatically. Their attraction is vastly weaker than their repulsion (by 40 or so orders of magnitude) so it is normally ignored.

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All bodies participate in the gravitational interaction, even a photon with a mass of 0 is gravitationally attracted. If the bodies have an electrical charge, then they also participate in electrical interaction. Therefore, your reasoning is correct, if on one side the bodies are pulled on the other side are repelled with equal forces and opposite in direction, then these bodies will be at rest.

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Of course. Gravitation is universal for all particles with mass, including charged particles. It is, however, very weak in comparison to electromagnetic interactions, so it usually is not very relevant in calculations (for particles).

For sufficiently heavy bodies with small amounts of charge, it is the other way around: gravitational forces are more relevant because the large mass offsets the weakness of the gravitational force.

Of course, if the heavy masses had a proportionate amount of charge on them compared to charged particles, electromagnetic forces would once again dominate, making gravitational forces irrelevant by comparison.

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Yes, they exert both gravitational and electrostatic forces on each other. And classically there is a possible neutral equilibrium.If you consider the particules in 1 dimension, separated by $r$, then the force between them is:

$$ \begin{align} F &= \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r^2} - G\frac{m_1m_2}{r^2}\\ &= \frac{1}{r^2} \left(\frac{1}{4\pi\epsilon_0} q_1 q_2 - Gm_1m_2\right) \end{align} $$

So $F = 0$ when

$$\frac{1}{4\pi\epsilon_0} q_1 q_2 = Gm_1m_2$$

And note that this is a neutral equilibrium: the net force is zero whatever $r$ is.

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Gravitation is universal. It applies to all particles having some mass including the charged particles. The charged particles produce both electric field and gravitational field and for both the fields the intensity is inversely proportional to the square of the distance. But due to the nature of the constants of proportionality used in the expressions of electric and gravitational fields (namely $G$ for gravitational field and $\frac{1}{4\pi\epsilon_0}$), the magnitude of the gravitational field is almost negligible concerning that of the electric field for normal masses around us. However, if the charged particles have very large masses but a small amount of charge on them, then the gravitational force between them may become comparable to or even larger than the electric force because the large mass balances the weakness of the gravitational force.

So for the last part of the question, yes it is possible from the above argument that the repulsive electric force gets balanced by the attractive gravitational force and the system remains in equilibrium.

I hope this helps you with the question.

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