I understand that both the electric and gravitational forces are inversely proportional to the squared distance from the point source and that the gravitational constant is around $10^{-20}$ times Coulomb's constant.  I don't understand why this suffices to conclude that the gravitational force is stronger than the electric force. The two constants that I see getting compared here have different units. One involves electric charges, while the other involves gravitational masses. To me, this makes as much sense as saying that a second is larger than a meter.  Is it related to the energy densities of these fields? If yes, how? I think that it is not the duplicate of the mentioned question because the point being asked in that question is why do we consider gravitational fields weaker even if they are more dominant in the universe at macroscopic levels. But, I want to ask that what are the criteria based on which we define the relative strength or weakness of forces/fields. What do we exactly mean by declaring a force/field stronger than the other? Is it related to their energy densities?
46238
I can understand that on small scales (within an atom/molecule), the other forces are much stronger, but on larger scales, it seems that gravity is a far stronger force; e.g. planets are held to the sun by gravity. So what does it mean to say that "gravity is the weakest of the forces" when in some cases, it seems far stronger?
75430
I usually think of gravitational potential energy as representing just what it sounds like:  the energy that we could potentially gain, using gravity.  However, the equation for it (derived by integrating Newton's law of gravitational force)...  $$PE_1 = -\frac{GMm}{r}$$  ..has me thrown for a loop, especially after .   If potential energy really meant what I thought it did, then it would always have to be non-negative... but this equation is always negative.  So what does "negative potential energy" mean!? If $KE + PE$ is always a constant, but PE is not only negative but becomes more negative as the particles attract, doesn't that mean the kinetic energy will become arbitrarily large?  Shouldn't this mean all particles increase to infinite KE before a collision? If we are near the surface of the earth, we can estimate PE as $$PE_2 = mgh$$ by treating Earth as a flat gravitational plane.  However, $h$ in this equation plays exactly the same role as $r$ in the first equation, doesn't it?   So why is $PE_1$ negative while $PE_2$ is positive?  Why does one increase with $h$ while the other increases inversely with $r$? Do they both represent the same "form" of energy?  Since $PE_2$ is just an approximation of $PE_1$, we should get nearly the same answer using either equation, if we were near Earth's surface and knew our distance to its center-of-mass.  However, the two equations give completely different answers!  What gives!?    Can anyone help clear up my confusion?
57203
Which of these laws is more fundamental or forms the basis of electrostatics? I started off with Coulomb's law and then I studied Gauss' law. I was wondering which one is more universal?  My professor derived Gauss' law using Coulomb's law but didn't do it the other way, so is Coulomb's law more fundamental? And can Gauss' law be used to prove the other?
14337
When I was in the university (in the late 90s, circa 1995) I was told there had been research investigating the $2$ (the square of distance) in the Newton's law of universal gravitation.  $$F=G\frac{m_1m_2}{r^2}.$$  Maybe a model like   $$F=G\frac{m_1m_2}{r^a}$$  with $a$ slightly different from $2$, let say $1.999$ or $2.001$, fits some experimental data better?  Is that really true? Or did I misunderstand something?
71566
Is there a difference between Weight and Force of Gravity or they just mean the same thing? My textbook does not clarifies this point.
63416
My knowledge on this particular field of physics is very sketchy, but I frequently hear of a theoretical "graviton", the quantum of the gravitational field.  So I guess most physicists' assumption is that gravity can be described by a QFT?  But I find this weird, because gravity seems so incredibly different from the other forces (yes, I know "weirdness" isn't any sort of scientific deduction principle).  For relative strengths:   Strong force: $10^{38}$ Electromagnetic force: $10^{36}$ Weak force: $10^{25}$ Gravity: $1$   Not only does gravity have a vastly weaker magnitude, it also has a very strange interaction with everything else.  Consider the Standard Model interactions:    No particle (or field) interacts directly with all other fields.  Heck, gluons only barely interact with the rest of them.  So why is it then that anything that has energy (e.g. everything that exists) also has a gravitational interaction?  Gravity seems unique in that all particles interact through it.  Then there's the whole issue of affecting spacetime.  As far as I'm aware, properties such as charge, spin, color, etc. don't affect spacetime (only the energy related to these properties).
53761
I understand that energy conservation is not a rule in general relativity, but I'd like to know under what circumstances it can still be possible. In other words, when is it possible to associate a potential energy to the gravitational field, so that the energy is constant in the evolution of the system?  Here are some examples, is there a convenient way to define energy in these scenarios?   Just a system of gravitational waves. A point mass moving in a static (but otherwise arbitrary) space-time. Equivalent (if I'm not mistaken) to a test mass moving in the field of a second much larger mass, the larger mass wouldn't move. Two rotating bodies of similar mass.   Overall, I'm trying to understand what keeps us from associating a potential energy to the metric. When we break the time translation symmetry of a system by introducing an electromagnetic field, we can still conserve energy by defining an electromagnetic potential energy. Why can't we do the same when we break TT symmetry by making space-time curved?
41591
I've been wondering, what causes the normal force to exist?  In class the teacher never actually explains it, he just says "It has to be there because something has to counter gravity."  While I understand this is true, it never explains why.  Whenever I ask anyone else they always respond in a similar way, saying "It has to be there, because the object is not accelerating", and this has become very frustrating.  So what is the cause of the normal force? From my reasoning, it has to be one of the four fundamental forces. (Gravity, electromagnetism, the weak force, or the strong force).  It would seam to me that electromagnetism would make to most sense (electrons in the outer shells of atoms repelling each other),   However, just as I thought this had to be right, I read a thing online about "certain fundamental particles repelling each other when their wave functions overlap".  I haven't studied quantum mechanics yet so I'm not really sure what to make of that.  If anyone could shed some light on this for me it would be much appreciated.
41605
Particles in gravitational fields are subject to gravitational time dilation. The closer a particle is near a gravitational source, the slower is running its clock. I would like to know more about the relation between gravity and gravitational time dilation.  In order to get a rough impression I used Newton's gravity equation (which may be used for weak fields, and I found that gravity and time dilation are (approximately) proportional: Can this result be confirmed on the base of Einstein's field equation (maybe even for stronger fields)?  dτ = proper time of a particle in the gravitation field of Earth, dt = proper time of an observer in infinity, rs = Schwarzschild radius of Earth, r = distance particle - center of Earth  Gravitational time dilation:  $\frac{dτ}{dt}=\sqrt{1- \frac{r_s}{r}} ≈ 1- \frac{r_s}{2r}$   Time dilation (difference):  $1-\frac{dτ}{dt} ≈ \frac{r_s}{2r}= \frac{GM}{c^2 r}$   Gravitational force (Newton's equation):  $F=G \frac{mM}{r^2}$  \begin{equation} \frac{Gravitational\:force}{Time\:dilation\:(difference)} ≈   \frac{G\frac{mM}{r^2}}{\frac{GM}{c^2 r}}=\frac{mc^2}{r}=\frac{rest\:energy\:(of\:the\:particle\:subject\:to\:time\:dilation)}{distance\:(of\:the\:particle)} \end{equation}  (As a result, time dilation would be approximately gravitaty, divided by the rest energy of the particle, multiplied by its distance.)
24614
Since Newton's law of gravitation can be gotten out of Einstein's field equatons as an approximation, I was wondering whether the same applies for the electromagnetic force being the exchange of photons. Is there an equation governing the force from the exchange of photons? Are there any links which would show how the Coulomb force comes out of the equations for photon exchange? I know that my question is somewhat similar to the one posted here , but it doesn't really have an answer to my question specifically.
14494
Consider the electric field surrounding an electron. The electron moves. A wave propagates at $c$ reflecting the electron's movement. This is an electromagnetic wave. Consider a mass. Classically surrounded by a gravitational field. The mass moves. The field deforms and it propagates at $c$.    Is this not a gravitational wave?  Why do people talk as if only general relativity predicted gravitational waves?
95828
I know this has been asked around, my question is a bit more specialized towards my personal problem.  If I have an object the magnitude of the weight is usually equal to the normal force, right?   I could be mistaken but from what I have read the normal force is the force the object is feeling opposite to the $(mass*gravity)$. If that is the case, and I put my hand on the object does my Force get added to the object's weight and thus the normal force of that object increases? Therefore if I keep adding hands to the object will the normal force keep compensating?   Also is the normal force the negative charged electrons repelling each other (electromagnetism)?
118043
Let's say that we want to measure the gravity force in 1D, 2D, 3D  and higher spatial dimensions.   Will we get the same force strength in the first 3 dimensions and then it will go up? How about if we do this with Electromagnetic force?   I've include the Electromagnetic force just to see if I can find an analogy to the gravity force behavior.
10406
Many sources state that the Earth's gravity is stronger at the poles than the equator for two reasons:   The centrifugal "force" cancels out the gravitational force minimally, more so at the equator than at the poles. The poles are closer to the center due to the equatorial bulge, and thus have a stronger gravitational field.   I understood the first point, but not the second one. Shouldn't the gravitational force at the equator be greater as there is more mass pulling the body perpendicular to the tangent (since there is more mass aligned along this axis)?
243
I am given two charged particles of same charge at a distance of $r$. They initially apply force $F$.  Now an infinite dielectric (of dielectric constant $4$) of width $\frac{r}{2}$ is introduced between the particles. What will be the new force?   I find this problem confusing because I have only been told about forces when its either fully dielectric or fully vaccum given by Coulomb's law. How do we get forces when only partial space is dielectric?
19392
Coulomb's Law states that the fall-off of the strength of the electrostatic force is inversely proportional to the distance squared of the charges. Gauss's law implies that the total flux through a surface completely enclosing a charge is proportional to the total amount of charge. If we imagine a two-dimensional world of people who knew Gauss's law, they would imagine a surface completely enclosing a charge as a flat circle around the charge.  Integrating the flux, they would find that the electrostatic force should be inversely proportional to the distance of the charges, if Gauss's law were true in a two-dimensional world. However, if they observed a $\frac{1}{r^2}$ fall-off, this implies a two-dimensional world is not all there is. Is this argument correct? Does the $\frac{1}{r^2}$ fall-off imply that there are only three spatial dimensions we live in? I want to make sure this is right before I tell this to my friends and they laugh at me.
20753
A line is charged uniformly with the linear density $\lambda$ . What is the force between two sections with lenghts $l_{1}$ and $l_2$ if their end points are separated with distance $r_{0}$?  I have calculated that the electric field from the section on the right is:  $$ E(x)=-k \lambda \frac{l_{2}}{(l_{1}+x)(l_{1}+l_{2}+x)}  $$  I am not sure how to get the force. I tried:  $$ dF=E(x)dq=\lambda E(x)dx \Rightarrow F=\int_{0}^{l_{1}}-k \lambda^2 \frac{l_{2}}{(l_{1}+x)(l_{1}+l_{2}+x)} dx $$ But probably it isn't the right way.   Thank you for any help in advance.    ------------------------------------------------------------------------------  propposed solution:      $$ dF(x_1)= (\lambda dx_1) \int_{l_{1}+r_{0}}^{l_{1}+r_{0}+l_{2}} \frac{\lambda dx_2 }{4\pi \epsilon_0 (x_2-x_1)^2}=k\lambda^2\frac{l_{2}}{(x_{1}-l_{1}-r_{0})(x_{1}-l_{1}-l_{2}-r_{0})}dx_1 $$ The net force:  $$ F=\int_{0}^{l_{1}} F dx_1=\int_{0}^{l_{1}}k\lambda^2\frac{l_{2}}{(x_{1}-l_{1}-r_{0})(x_{1}-l_{1}-l_{2}-r_{0})}dx_1=k\lambda^2log(\frac{(l_{1}+r_0)(l_2+r_0)}{r_0(l_1+l_2+r_0)}) $$