why is electric force related to mass?

We know a mass attracts mass with a force $F=G\dfrac{mM}{r^{2}}$ and the acceleration due to that force is $a=G\dfrac{M}{r^{2}}$ where $F=ma$.

Now for charges, what observations prevent us from defining $F=qa$ where $F=k\dfrac{qQ}{r^{2}}$ and acceleration due to the force is $a=k\dfrac{Q}{r^{2}}$.

Why can't we say the charge $q$ which is under the influence of a force is moving because of its charge and not by its mass.

Edit-

I actually meant to ask "why the movement of the charge is governed by interia and F=ma"

• "Why can't we say the charge q which is under the influence of a force is moving because of its charge and not by its mass" But... It is moving because of it's charge. Just like the formula shows. "why is electric force related to mass?" How do you mean related? Where is that relation shown in the question? I'm not sure I see the issue you are asking about. Would you mind clearing it out for me? – Steeven Jan 29 '17 at 9:45
• I actually meant to ask "why the movement of the charge is governed by interia and $F=ma$" – N.G.Tyson Jan 29 '17 at 9:49
• You should read about mass spectrometry. That is one situation where the charge to mass ratio becomes important. – Raziman T V Jan 29 '17 at 9:50
• Its unclear what you are asking ,can you please edit the question? – Lapmid Jan 29 '17 at 10:05
• Aha! I see. If you will edit that into your question, the whole thing will make more sense and I'll remove my downvote. – Steeven Jan 29 '17 at 10:25

I actually meant to ask "why the movement of the charge is governed by interia and F=ma"

Simple:(You can do the following experiment even in your home)

Take two balls $A$ and $B$ each of charge 1 coulomb and each of mass 1 kg. Now place them at a distance of 1 metre.

Take another two balls $A'$ and $B'$ each of charge 1 coulomb. But mass of ball $A'$ is 1 kg and mass of ball $B'$ is 2 kg. Now place them at a distance of 1 metre.

In both cases, forces are same (by Coulomb's law) because the charges on each ball and the distance between them are same.

However since the mass of balls $B$ and $B'$ are different in both cases, accelerations of balls $B$ and $B'$ would be different. It tells us that the accelation is dependent on mass.

• How do I find a ball with a charge of 1C? – Gonenc Oct 21 '17 at 13:33

It is somewhat of a coincidence that inert mass and gravitational mass is the same thing. Just from the equations, the $m$ in $F = ma$ and $F = G mM/r^2$ would not have to be the same. If it was different by some factor however, one could just redefine $G$ to make the two $m$ the same again.

Electric fields act on a different charge. Gravity acts on the “gravitational charge” $m$ whereas the electric field acts on the electric charge $q$. This factor $k$ (which can be $1/4 \pi \varepsilon_0$, depending on the unit system) gives the coupling strength from the electric field to that charge.

The actual movement of the charge is also governed by interia, therefore you also need $F = ma$. In general we do not have $m \propto q$ as we had with gravity. We cannot redefine $k$ in such a way that $m \triangleq q$ would work out. The problem is that charge and mass are independent of each other, particles with virtually the same mass (neutron vs. proton or up-quark vs. down-quark) have vastly different electric charges ($0$ vs. $+1$ or $+2/3$ vs $-1/3$). You can redefine $G$ only once, and then there have to be conversion factors for the other forces.

What one can do, however, is to define the force field $\vec E$ such that you have something similar looking for both: $$\vec F = m \vec a \,, \qquad \vec F = q \vec E \,.$$

Because 'mass' measures the degree in which matter resists a change in velocity. Unsurprisingly, it is proportional to the amount of material (the more material is there, the harder it is to move it).

Now, sometimes you see "fictitious forces" - like the centrifugal force, which yield the same acceleration independent of mass. This is because those are not really external forces, but rather an outcome of forgetting to account the acceleration of your view point (as in sitting on the surface of a rotating sphere). This is called "non-inertial frame of reference" by physicists.

The strange fact is that Gravity itself seems like a 'fictitious force'. Einstein's relativity theory emanates from that, with the claim that Gravity is indeed a feature of the structure of curved space-time, and not really a 'Force'.

I actually meant to ask "why the movement of the charge is governed by interia and F=ma"

In classical mechanics it does not matter what form the potential takes, and you end up with a force law that simply says $F=m \frac{d^2v}{dt^2}=ma$.

So the potential (which produces that electrostatic force) is just treated the same as any potential (or force) in classical mechanics.

I suspect you are being thrown off by considering gravitational force, which is proportional to the mass of both objects, so that acceleration of an object has no dependence on it's mass. In general forces don't necessarily depend on the mass of objects, so the object's mass won't generally conveniently cancel like that.

The question is whether the answers are true where it is really important, in General Relativity, and not just Newtonian physics.

The answers provided so far are correct. Electric charge does not couple directly to the gravitational field, but rather to the electromagnetic field. The equivalence principle in essence says that the source of the gravitational field, the gravitational mass, is proportional to the inertial mass, and thus one cannot distinguish uniform gravitational fields from accelerated frames. The same is not true or electric charge, it is not proportional to inertial mass (it is an independent property of a particle instead), and thus acceleration will actually depend on q/m, the charge to mass ratio.

This explains the a argument for classical Newtonian physics, where the equation F = ma comes from and is true. But there is no such equation in General Relativity (GR). So the question is, do we know that the same is true in GR? We know it should be, that GR was constructed to obey the equivalence principle for gravitational fields. How were other forces (or fields) accounted for, and is motion again dependent on q/m?

The answer is yes, geodesic motion in GR with an electromagnetic field which contributes to the motion, for a charged particle, just depends on q/m.

The way that comes about is how Maxwell's equations, and the equations of motion, were included in the so called Einstein Maxwell equations. The Maxwell's equations were included in the Einstein equations as part of the right hand side, the stress energy tensor, and the covariant Maxwell's equations.

The Einstein Maxwell equations, and the covariant Maxwell's equations ('simply' replace commas by semicolons, i.e., make derivatives covariant) are in

https://en.m.wikipedia.org/wiki/Einstein_field_equations#Einstein.E2.80.93Maxwell_equations

The right hand side is how the electromagnetic field contributes to the gravitational field. The spacetime is called an electrovac spacetime

The next paper derives and describes, in detail, the geodesics in the resulting electrovac spacetime produced by charged black holes (they did it for static, rather than stationary spacetimes, but it's been done for stationary Kerr Newman spacetimes also).

https://arxiv.org/pdf/1011.5399.pdf

It describes test particle motion in Einstein Maxwell spacetimes. Depends on particle charge (and theoretically magnetic charge, if it existed). Mass gets normalized out, i.e. It's the ratios of q/m and theoretically $q_m$/m, the magnetic charge to mass ratio, if magnetic charge exists, as well as the metric, that determines the motion of the particles.

So, yes, it is built into 4D GR.

It is interesting that in 5D, as is well known, one can actually get an equivalence principle. In 4D it is instead q/m, as in classical physics.

$F=ma$ has nothing to do with $F = G\frac{m_1m_2}{r^2}$ nor does it have anything to do with $F = k\frac{q_1q_2}{r^2}$.

You arrive at $F = ma$ from Newton's second law.

You arrive at $F = G\frac{m_1m_2}{r^2}$ from Newton's Gravitational Law.

You arrive at $F = k\frac{q_1q_2}{r^2}$ from Coulomb's Law.

You are asking this question because you thought $F = ma$ and $F = G\frac{m_1m_2}{r^2}$ were related. However, there are formulae which relate a field quantity to a force:

$F = mG_{field}$

$F = qE_{field}$

In electrostatics charge is moving under the influence of an electric field $F=qE$ F and E are vectors if we put $F=qa$ -we wouldn't be able to know the direction of the force but it is possible to know the acceleration of the charge by applying Newton's second law $F=ma=qE$ therefore $a=qE/m$ so there is an influence of mass there