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The formula for electric force is given by $F = k \dfrac{q_1q_2}{r^2}$ Where $q_1$ and $q_2$ are the charges, and $r$ is the distance. Therefore, there is no lower limit. It is always given by this formula.


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If the charges are kept at a fixed distance R, the force will be given by: $$ F = \frac{kQ_1Q_2}{r^2} $$ The smallest possible charge that can exist freely is that of an electron or proton which is numerically equal to $1.6 \times 10^{-19}$ coulomb, put the values in the equation and you get the answer.


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There is no range of electric force values, F = k*q/r^2 It will be fixed unless you having a mechanism that makes the charges to vary as required.


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In case the question concerned the case $r \rightarrow 0$, you would reach the situation where the charge (represented by a charged particle like electron, proton, positron) approaches the Coulomb field of the other particle and they would have a tendency to create a kind of a planetary system - but - quantum effects start to play a role here and those two ...


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If r = 0 then you have a single charge, so the problem reduces to the electromagnetic self-force problem. A charge will interact with the electric field it is in, and that includes the field due to its own charge.As long as the charge is not accelerating, one can pretend as if there is no self-force, but for accelerating charges, the self-force will lead to ...


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The reason being the charge imbalance in the frame of the moving electrons would produce a smaller charge imbalance force Since $$k = \frac{1}{4\pi \epsilon_0}$$ decreasing $k$ is equivalent to increasing permittivity $\epsilon_0$ of free space. Assuming the permeability $\mu_0$ of free space is held constant, decreasing $k$ decreases the invariant ...


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What is wrong with my reasoning? Opposite charges attract because one of the charges has a negative sign. The force on the negatively charged particle is thus $$\vec F_- = \frac{kQ(-Q)}{r^2}\hat r = -Q\,\frac{kQ}{r^2}\hat r = -Q\,\vec E_+ $$ The force on the negatively charged particle is opposite the direction of the field from the positively ...


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The electric field lines show the direction of the electric force acting on a unit positive charge at a particular point in space. So, therefore, the force acting on a negative charge, as is in your question, will act in the opposite direction shown by the electric field lines. I believe that the fact that field lines are defined in terms of a unit positive ...


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A neutral object can be induced a non-zero charge when placed in an electric field. The charges or dipoles within that material will simply rearrange or rotate to aline slightly. An electric field will be generated, which will counteract the current field. Have a look at dielectrics. The gravitational constant $G$ is... A constant. Just like the ...


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The electron is relatively close to the uniformly charged sheet. You can simplify the problem to a electron in the homogeneous field of a capacitor consisting of two square plates with surface $A=0.25 m^2$. The force to a electron $e$ between the plates of the capacitor is $F=eE$, where $E = \frac{Q}{\varepsilon_0 \varepsilon_r A}$ is the strength of the ...


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The metal plate, being a good conductor, will have its electrons rearrange in such a way as to neutralise the electric field inside the plate. The electrons would tend to bunch up in the plate at the point(s) closest to each of the two charges, alterring the electric field that they're exposed to and changing the electrostatic force on them. Strictly ...


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will this cause a change in the electrostatic force acting on one charge due to another? ans is no. The two charges will induce some charges on the metal plate which will ofcourse change the electric field but according to principle of superposition force on one charge due to another charge is not affected by the presence some others charges near.(I meant ...


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Strictly talking about electrostatic conditions(not referring to induction): Coulomb's law has a constant in it Ke(k electrostatic) which is equal to 1/4πε where ε is the electric permittivity of the medium that the electric field is in.More electric flux exists in a medium with a low permittivity. So,this is the difference in your equation between the two ...


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The Coulomb "interaction" appears as the answer to a very specific question, even in QED, that is "what is the correction to the ground state energy of the electromagnetic field if two charges $q_1$ and $q_2$ are pinned at specific locations separated by a distance $r_{12}$?" . The answer to that question is exactly $\Delta E_0 = \frac{q_1q_2}{4 \pi ...


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Your brain/mind might be processing the information in the wrong way. As field is uniform, force remains constant and acceleration remains constant. You see, the acceleration remains constant, but velocity doesn't remain constant, as the time to which your proton is accelerated increases, your proton's velocity also increases. So, your proton will be ...


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NO the force does not change with distance as E=F/Q F=QE electric field is constant and charge is also constant then force is also constant.



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