I was calculating potential energy due to point charge at origin using relationship $$U=\dfrac{1}{2}\epsilon_{0}\displaystyle\int \mathbf {\vec E.\vec E\ }  \ d\tau$$ and it comes out to be infinity  because $\mathbf {\vec E.\vec E\ }=|\vec E|^2 $ falls off as $\propto$ $\dfrac{1}{r^4}$ and volume $d\tau$ grows as $4\pi r^2 dr$ resulting in overall fall off as $\propto$ $\dfrac{dr}{r^2}$ and which in turn lead to blowing up of integral at $r=0$ and U approaching $\infty$ Question: it is written in my textbook as it is that: &quot;it takes infinite energy   to pack a point charge(like electrons, protons) into zero volume.&quot; i want to know physical justification of this result (not mathematical derivation like above )
124240
I will try to limit the question in the case of the electric fields, but is something that applies also to the magnetic ones. There are two ways to express the energy in a capacitor:    By Voltage : $U = 1/2 CV^2 $ And by Field : $U = 1/2 \varepsilon E^2Ad$, With Energy Density: $u = 1/2 \varepsilon E^2 $    Unless i understood everything wrong and these two are NOT the same quantity, i have the following question. When we have two charges placed at points A and B, then in order to calculate the energy of the system, we will take the first charge, place it at point A WITHOUT doing any work, and then we will calculate the work needed to place the second charge at point B.  The weird thing to me here, is that while we have placed the first charge, without generating any work the system will still have the energy held in the field of the charge!   There is obviously something that i miss, but what?
94763
I am trying to derive the formula for the self-energy of a non-conducting sphere. I know it to be $$E = \frac{3Q^2}{40\pi\epsilon_0 R}$$  I am trying to use the energy density formula for it which is  $$ \frac{dU}{dV} = \frac{1}{2}\epsilon_0 E^2$$  My method was to find the general formula for the $E$-Field inside the non-conducting sphere, which is $$ E = \frac{Qr}{4\pi\epsilon_0R}$$ Then using that, and setting $dV=4\pi r^2dr$, then solving for U using the energy density formula, by integrating between 0 and R (the radius of the sphere). However, in doing this, I am out by a factor of 3, and cant figure out why. Any ideas? Or have I just messed up somewhere, I have looked extensively through my written work and everything seems sound, so I suspect the issue is with my physical intuition.
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Suppose I have a sphere of radius $r$ with all the charge residing on the surface, distributed  uniformly i.e. charge density $\sigma$ is constant. I want to find the electric field created by this sphere.  All problems of this sort I have solved by enclosing the sphere in a larger imaginary sphere with radius $R$, and then using the assumption that the electric field lines are in a radial direction, in combination with the divergence theorem.   If I use that method for this question, I find that the Electric field inside the sphere is zero. I can't understand this intuitively. I would have thought that if you are in the proximity of charge, there will be an electric field. Why would the electric field lines extend outwards radially from the sphere, and not inwards too?
33158
A solid sphere has charge $q$ and radius $R$. Find the potential at a point a distance $r$ from the center of the sphere where $r&gt;R$, using infinity as the reference point.  My attempt:  From Gauss' theorem we may deduce that $\displaystyle\mathbf{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r'^2}\hat{\mathbf{r}}$ where $r'$ is the distance of an arbitrary point from the center of the sphere provided $r'&gt;R$.  $V=-\int_\infty^\mathbf{r}\mathbf{E}\cdot d\mathbf{l}$  My question: What is $d\mathbf{l}$? Since we are traversing in the direction opposite to $\hat{\mathbf{r}}$, I think it should be $-dr'\hat{\mathbf{r}}$. But when I use it to find $V$, I get a sign error. Please help!
22298
I just began studying electrostatics in university, and I didn't understand completely why the electric potential due to a conducting sphere is  $$   V(\vec{r})=\begin{cases}     \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R}, &amp; \text{if $r \le R$}.\\ \\     \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r}, &amp; \text{if $r \gt R$}.   \end{cases} $$  Where Q is the total charge and R is the radius of the sphere (the sphere is located at the origin).  I only understand the second part of this equation (when $r &gt; R$).  I know Gauss Law. know the charges go to the surface. I know the electric field strictly inside it must be zero. (I also know the electric field is not defined for a point that lies exactly in the surface). And I know $\vec{E} = -\nabla{V}$.  Therefore, I know the electric potiential inside the sphere must be constant. Let $C$ be this constant.  It seems that  $$ C = \lim_{r \to R^+} V(r) = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R} $$  But why is this true? My textbook says: because the electric potential must be a continuous function. But why? I am hoping for a non-experimental reason.  Please be precise when mentioning $r&lt;R$ or $r\le R$. Those are different and I get easily confused when people misuse those.
114044
I have come to know the electrostatic potential energy in vacuum is given by $${\frac{1}{2}} \epsilon_0\int d^3x {E^2} $$ and this energy  is due to the mutual electrostatic coulomb potential energy. So only a single point charge in space also contributes to some electrostatic potential energy according to $${\frac{1}{2}} \epsilon_0 \int d^3x {E^2} $$  as due to a single point charge also we have an electric field at every point in space. Now I am not getting it physically, why should there be any Coulomb energy in this case as there is no other charge to provide the electrostatic coulomb potential energy?
105813
So in my textbook, it states that the $E=hf$ equation applies to electrons, and all particles, not just photons.   But in order to prove this, wouldn't these particles need to have zero mass, to satisfy $E=pc$ from Einstein's equation?  In Einstein's theory of relativity $E^2= p^2c^2 + m^2c^4$, we get $E=pc$ because we consider the mass to be zero.   For $E=hf$ to work for other particles, don't we need to assume electrons have zero rest mass?
106610
My question is loaded with assumptions so to minimize them, I would like to ask it with respect to an electron and anti-electron annihilating.  When I think of annihilation, I think of electron and positron turning into energy where the total energy represents the combined masses plus kinetic energy each particle had prior to the collision and annihilation (Assumption 1).  But, doesn't the field of each charged particle also have an energy content and does this energy content of the charged particle field show up in the final energy?  I have never seen this though, only the sum of the mass energy. (Assumption 2 possibly).  Also, how is it that the masses annihilate if they are not ANTI to each other.  Mass has no opposite like charge does (or other things).  Does the annihilation of the charges also somehow annihilate the masses and if this is the case, that would require energy so this energy should show up somewhere.  Or is there some other opposite "thing" that results in annihilation of mass?  Physicists used to think that electron mass was actually the inertia of moving charge in a field (or something like that).  Is there some kind of connection between mass and charge such that charge annihilation is also mass annihilation?
89382
In the  (semi-classical, but should still apply here I think), the conducting electrons are in a constant electric field, and, in between collisions with the lattice ions (that happen, on average, in a time $\tau$ since their last collision) in which their velocities are reset to an average of 0, they are accelerated and gain a momentum  $$p = qE\tau (=mv)$$, where $q$ is its charge and $E$ is the electric field from the applied voltage. This obviously means in between the collisions, the electron accelerates $a$, from $v_f = v_i + a\tau$.  Accelerating charges give off . So does a DC carrying wire give off radiation one can measure?  I suspect it does, but it's of an insanely small magnitude. A quick order of magnitude calculation with the following values gives me:  $$q \sim 10^{-19} \,{\rm C}, \\ E\sim 1\,{\rm V}/1\,{\rm cm}=100\,{\rm V/meter},\\ m\sim10^{-30}\,{\rm kg} \\ \Rightarrow a = qE/m \sim 10^{13}{\rm m/s}^2 \\ \Rightarrow P \sim q^2a^2/\epsilon_0c^3 \sim 10^{-28}\,{\rm W}$$  Which is obviously pretty small. I'm not sure what's the smallest power that can be measured, but I'm guessing by increasing the applied voltage one could increase the radiated power. But is the idea even fundamentally sound?
27131
There is some confidence that electron is a perfect point e.g. to simplify QFT calculations. However, searching for experimental evidence (),  only points argument based on g-factor being close to 2:  extrapolating from proton and triton behavior that  for particles composed of 3 fermions should be $\approx g-2$:    Using more than two points for fitting this parabola it wouldn't look so great, e.g. neutron (udd) has $g\approx-3.8$ and .  And while classically $g$-factor is said to be 1 for rotating object, it is for assuming equal mass and charge density ($\rho_m\propto\rho_q$). Generally  by modifying charge-mass distribution:  $$g=\frac{2m}{q} \frac{\mu}{L}=\frac{2m}{q} \frac{\int AdI}{\omega I}=\frac{2m}{q} \frac{\int \pi r^2 \rho_q(r)\frac{\omega}{2\pi} dr}{\omega I}= \frac{m}{q}\frac{\int \rho_q(r) r^2 dr}{\int \rho_m(r) r^2 dr}$$  Another argument for point nature of electron is tiny cross-section, so  for electron-positron collisions:    Beside some bumps corresponding to resonances, we see a linear trend in this log-log plot: $\approx 10^{-6}$ mb for 10GeVs (5GeV per lepton), $\approx 10^{-4}$ mb for 1GeV. The 1GeV case means $\gamma\approx 1000$, which is also in Lorentz contraction: geometrically means $\gamma$ times reduction of size, hence $\gamma^2$ times reduction of cross-section - exactly as in this line on log-log scale plot.   More proper explanation is that it is for collision - transforming to frame of reference where one particle rests, we get $\gamma\to\approx \gamma^2$. This asymptotic $\sigma \propto 1/E^2$ behavior in colliders is well known () - wanting size of resting electron, we need to take it from GeVs to E=511keVs.  Extrapolating this line (no resonances) to resting electron ($\gamma=1$), we get $\approx 100$ mb, corresponding to $\approx 2$ fm radius.  From the other side we know that two EM photons having 2 x 511keV energy can create electron-positron pair, hence energy conservation doesn't allow electric field of electron to exceed 511keV energy, what  from $E\propto 1/r^2$:   $$\int_{1.4fm}^\infty \frac{1}{2} |E|^2 4\pi r^2 dr\approx 511keV$$  Could anybody elaborate on concluding upper bound for electron radius from g-factor itself, or point different experimental boundary?  Does it forbid electron's parton structure: being "composed of three smaller fermions" as Dehmelt writes? Does it also forbid some deformation/regularization of electric field to a finite energy?
97164
I'm currently reviewing electricity for the AP Physics C exam, and came across a problem or general topic I don't really understand that has to do with Gauss's law.  The general form of Gauss's law we learn in AP Physics C is $$\int E \;\mathrm{dA} = q_{enclosed}/ε_0$$  So when the enclosed charge is 0, as in the case of a hollow spherical shell with uniform charge distribution, it follows that the electric field is zero in the sphere.  However, when the charge distribution becomes non-uniform, the enclosed charge is still zero, implying that the electric field is still zero. However, if all the charge on the sphere is moved to one side, there should be an electric field directed from one hemisphere to the other. I'm wondering how this application of Gauss's law here is wrong and why.  When writing up this question I found this link:  where someone answered a similar question. However, I still don't understand why if flux is zero it does not mean the electric field is zero. Please keep in mind that I only have taken Calculus up to AP Calculus BC, which I think roughly corresponds to Calculus II.
40103
As written in Principles of Physics by Halliday, Resnick and Walker:     The electric field is a vector field; it consists of a distribution of vectors,one for each point in the region around a charged object. In principle, we can define the electric field at some point near the charged object. We first place a positive charge $q_0$, called a test charge,at the point. We then measure electrostatic force $\vec{F}$ that acts on the test charge. Finally, we define the electric field $\vec{E}$ at that point due to the charged object as $$\vec{E} = \frac{\vec{F}}{q_0}$$ .   So, this is the way they derive the electric field. But one thing that perplexes me is that as test charge has its own field, so when the test charge is introduced in the field of concerned charged object, say another point charge $+q$ , won't it get shifted due to the repulsion from the test charge $q_0$ (which is, by convention, positive)? Then consequently the field of $+q$ will also be shifted. In a word, it will get disturbed by that test charge. But the book derived the formula as if the positive point charge $+q$ remained fixed i.e. the test charge had no effect on the field of the point charge. But how can it be? If the test charge $q_0$ is a charge, then obviously it has also field which will exert force on $+q$. So how can the formula be true? Why did the book neglect this?  This phenomenon can be compared to earth's gravitational field. When a body moves upward, it exerts force on the earth downward. But as the earth has huge mass, its shifting can be neglected and hence it doesn't affect its gravitational field.   But here it is a point positive charge; it has no huge mass. So why did the book neglect its shifting and how can the formula be true? Please help.
41379
For example, in a dipole, at the center of the two charges making up the dipole, the potential is zero but the electric field is non-zero. But if $E = -\operatorname{grad}V$, then why is $E$ not zero?  Another example: In the case of two positive charges, the electric field is zero at the mid-point of line joining them, but potential is non-zero. What am I misunderstanding here?
67855
I have been thinking about ways of teaching electronics and I'm wondering if the following is true...  For starters, when we talk about voltage as energy per unit charge, is this energy manifest simply as the kinetic energy of the electron?  So, all other things being equal, will the electrons coming out of a 4 volt battery will have twice the velocity of electrons coming out of a 1 volt battery because of ${1 \over 2}mv^2$?  This would seem to imply that the voltage drop across different parts of the circuit is essentially a comparison of the electron speeds at those points.  Furthermore doesn't this imply that the electrons are nearly halted by the time they reach the positive terminal of the battery? This makes intuitive sense to me, it seems pedagogically sound, and it provides an explanation for how electrons "know" to give up their energy across the circuit -- electrostatic repulsion communicates later resistances to earlier parts of the circuit.  How much of the above is correct? The one things that makes me suspicious is that, in this picture, electron density at the positive terminal of the battery is very high, and repulsive forces would prohibit this. On the other hand, if the Joules per Coulomb are not coming from kinetic energy, then what? The field? But wouldn't that just dissipate as light?  Something is wrong but I don't know what.
76877
How can it be that, if like charges repel, they don't repel themselves? In other words, why don't charges break apart?  About the possible duplicate: I want to know about charges in general, not just that of an electron. My response to :  Thank you very much for the extensive explanation. Unfortunately it's a bit above my level (I'm a first year electrical engineering bachelor student). This is how I understand it: If the charge is the sum of multiple separate charges, there has to be an external force that keeps this charges together. However we most often think of electrons and protons as point charges. In other words, we don't think of them as if they are made up from different (smaller) parts. But the idea of electrons and protons as point charges has its own problems. I didn't know that the electric field has a mass. From the formula, I understand that $m\propto \frac{1}{r}$. This would mean that $m\rightarrow \infty$ as $r \rightarrow 0$ (or $r = 0$), which of course is physically not possible. But there is a way to calculate the radius of an electron based on the mass, the speed of light and the constant of Planck. (Unfortunately I don't know why.) You can get around this by a technique called called renormalization, which causes the integral to converge. Unfortunately, I don't understand the rest of the answer due to my level in physics. Nevertheless I am grateful for you answer. It would be awesome if you could confirm or correct my understanding.
45877
If a capacitor is charged by a source with constant voltage, $V$, the energy, $W$, stored by a capacitor of capacitance $C$ is given by:  $$ W = \frac{1}{2} QV = \frac{1}{2} V^{2}C = \frac{Q^{2}}{2C} $$  where $Q$ is the charge stored on the plates.  If the capacitor is now disconnected, and its capacitance is changed by increasing the distance between the plates, its energy must increase since work is being done by moving charge under the influence of an electrical field. Furthermore, since the plates are not connected to anything, it can be assumed that charge is conserved.  The method for calculating the change in energy that I have seen in textbooks and on the internet is to calculate $Q$ for the pre-separated case, calculate $C$ for both cases, then use:  $$ W = \frac{Q^{2}}{2C} $$  to give $$ \delta W = \frac{Q^{2}}{2} \left( \frac{1}{C_{2}} - \frac{1}{C_{1}} \right) $$  where $C_{2}$ is capacitance post-separation, $C_{1}$ is capacitance pre-separation and $\delta W$ is the change in energy.  This seems naive to me. That formula was for the process of charging a capacitor from a constant voltage with the $\frac{1}{2}$ arising from the potential difference that charge is raised by decreasing as more charge is deposited.  With constant $Q$, we can say that:  $$ dW = QdV $$  and integrate to give:  $$ \delta W = Q(V_{2} - V_{1}) $$  where $V_{2}$ is the post-separation voltage and $V_{1}$ is the pre-separation voltage. Substituting $Q = CV$ then gives:  $$ \delta W = Q^{2} \left( \frac{1}{C_{2}} - \frac{1}{C_{1}} \right) $$  i.e. double the textbook answer.  Am I missing something?
70100
I know how to calculate the kinetic energy of the emitted electrons   $$KE=hf-\phi$$  And the number of the emitted electrons, but this is valid for a certain frequency  $$N=\frac{IA}{hf}=\frac{P}{hf}$$  What happens if I have a frequency range? for example between λ1 and λ2?  BTW: It's not homework, I'm an engineering student,  not a physics student