Be my hero and explain this to me! How does the electric force change when the amount of one charge is doubled? Thanks :D

Question

anonymous · Answer

Look at Coulomb's law, you will see that the force between two charges is proportional to the size of each charge

anonymous · Answer

so, If one of the charges is doubled, then the Eforce gets larger?

anonymous · Answer

you can be more specific than that, if one charge is doubled, then the force is ...... ?

anonymous · Answer

Could you give an example?

anonymous · Answer

just look up coulomb's law

anonymous · Answer

I don't believe the question calls for anything more specific than that, but just so I can understand it better I will.

owlcoffee · Answer

Coulombs law states that the electric force of two charges is given by the following formula:
$$F_e = K \frac{ qQ }{ r^2 }$$
where "k" is the constant of coulomb, "q" is a given charge ang "Q" another static charge,
and "r" being the distance between the two charges.
at the same time "k", the constant is:
$$K=\frac{ 1 }{ 4 \pi \epsilon_0 }$$
"epsilon zero" is the relative permitivity, but doing some maths we get:
$$K=8.987\times 10^9$$

Now, let's see what happens if I put two charges in space and calculate the force, say they are 2m from each other, and q is 1coulomb and the other 2 coulombs:
$$F_e =K \frac{ 1.2 }{ 2^2 }$$
$$F_e= 4.49 \times 10 ^9 N$$
Now, what would happen if the second one, was 4 Coulombs?
$$F_e= K \frac{ 1.4 }{ 4 }$$
$$F_e = K \frac{ 4 }{ 4 }$$
$$F_e= 8.98 \times 10 ^9N$$
We can observe that the range of error is minimal, but we can be sure that the force doubled, wich is obvious because the force is directly proportional to the product of the charges, so if it doubles, the force doubles as well.