If you have two point objects both the same positive charge and both of the same mass at a distance $r$ from each other.

The force between them due to gravity is $F_g=\frac{Gmm}{r^2}$ and $F_g$ is also positive.

If the objects are not charged at all, then they will move towards each other as the force is positive.

But as they are charged the force between them due to the electric charge will be: $F_e = \frac{QQ}{4\pi\epsilon_0r^2}$ And $F_e$ is positive so if you take the resultant force on the objects you get $F_g + F_e$ as they are both positive they're both the same direction and the particles will attract with a greater acceleration than if they werent charged

Say $|{F_e}| = |F_g| = F$ then the force between the two particle would be $2F$ Except that's not the case as the charges are both the same so $F_e$ is repulsive and as its the same magnitude as $F_g$, the particles will not move as the forces cancel each other out.

Basically my question is, how can the signs be the same for both equations but the actual directions of the forces be different?

  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$
    – Qmechanic
    Jan 13, 2013 at 19:29

1 Answer 1


It's just convention. The equations are written down to describe phenomena, and the notion is set up in such a way that we understand what the directions mean. If it makes you feel more secure, we could write down $F_{e} = k\frac{q_{1}q_{2}}{r^{2}}{\hat r}$ and $F_{g} = -G\frac{m_{1}m_{2}}{r^{2}}{\hat r}$, but there's no need to, because we all know that gravity is always attractive and that opposite charges attract and that alike charges repel.


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