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1

That is a unusual case but it can be achieved in a particle accelerator. when they give electrons enough kinetic energy to overcome the electric force they have on each other. keep in mind that electrons, small as they are, do have size. so for them to "touch" you will need A LOT of energy, but not infinite.


0

Suppose we have in static electricity two charges negative and positive equal in absolute value and we put electrons along the road between them ,we know that the electrical force or field is different from point to point between them because of distance according to columb law and since the force is different then every electron have different ...


7

There is another 'infinity' (among others) lurking in classical electrodynamics which is evident when one calculates the electrostatic energy $W$ of a uniform spherical charge distribution of radius $a$ and total charge $Q$ $$W = \frac{3}{5}\frac{Q^2}{4\pi \epsilon_0 a}$$ Thus, by this result, a point (zero radius) particle of charge Q has 'infinite' ...


11

You are correct when you concluded that two classical point electrons could never touch each other. It would take infinite energy.


1

The modern mind picture of electrons bound to a nucleus is that of an orbital rather than of an orbit as you seem to be thinking. However, your ideas are still somewhat meaningful in this modern picture in that region of electron delocalisation is more tightly confined around the nucleus for lower energy orbitals than it is for higher energy ones. You might ...


0

I suspected that one needed to go back to the definition of the currents and indeed, in doing so one can derive the result. Here's a short version. The electron current is defined as [see equation (6.6) in 1] $$\tag{1}J_\mu(x) = -e\bar{u}_f\gamma_\mu u_i \times\mathrm{exp}[(p_f-p_i)\cdot x]$$ which we write as $$\tag{2}J_\mu(x) = ...


0

It seems to me that there is a typo in the book. Your starting equation should be the following: \begin{equation} T_{fi} = -i\int \frac{d\omega \,d^{3}\textbf{q}}{(2\pi)^{4}} \tilde{A}(\textbf{q},\omega)\tilde{B}(-\textbf{q},-\omega) \frac{1}{|\textbf{q}|^{2}}, \end{equation} where $\tilde{A}$ denotes the Fourier transform of $A$. Then, using ...


0

There are at least two (equivalent) ways: 1) Net surface charge would be induced at the interface between these two media as well as in the vicinities of the two point charges. You can then apply Coulomb's law to one of the point charge with the contributions from all other point and surface charges. This method requires integration. 2) You can use the ...


1

I think a diagram will help: So at the end, the resultant force on q1 is just the y components of each force exerted by the q2 and q3 charges added together


1

By symmetry, you need only consider the force on one of the charges on the corner, say the top right, due to the other four. There will be a repulsion from the other corners pushing it away from the centre, so you must pick Q to balance this with an attractive force of just the right size. One way to work this out is to calculate the vector forces ...


0

The "traditional" form of Coulomb's law, explicitly the force between two point charges. To establish a similar relationship, you can use the integral form for a continuous charge distribution and calculate the field strength at a given point. In the case of moving charges, we are in presence of a current, which generates magnetic effects that in turn exert ...


0

I found the answer to be .756N by multiplying .63N and sin(36.87).


0

Here is a big hint... I have to believe you can finish it from here. The key is "similar triangles".



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