$\vec{E}$ is electric field $r$ is distance between source and field points $\hat{r}$ is a unit vector from source point to field point $x,y,z$ are Cartesian coordinates of field point $x',y',z'$ are Cartesian coordinates of source points $r= \left[       (x-x')^2 + (y-y')^2 + (z-z')^2         \right]^{1/2} $ $\rho$ is uniform charge distribution density  \begin{align} \nabla \cdot \dfrac{(\hat{r})}{r^2} &amp;=\nabla \cdot  \left[     \dfrac{x-x'}{r^3} (\hat{i}) + \dfrac{y-y'}{r^3}(\hat{j})  + \dfrac{z-z'}{r^3}(\hat{k})    \right]\\ &amp;=\dfrac{\partial}{\partial x} \left(  \dfrac {x-x'}{r^3} \right)  +\dfrac{\partial}{\partial y} \left(  \dfrac {y-y'}{r^3} \right) +\dfrac{\partial}{\partial z} \left(  \dfrac {z-z'}{r^3} \right)\\ &amp;=0 \tag1\\ \end{align} $\text{(after some calculation, it will come out to be zero)}$ Using equation (1): $$\nabla \cdot \vec{E} =\nabla \cdot     \left[    k\ \rho \displaystyle\int_V  \dfrac{dV}{r^2}  (\hat{r})  \right] =k\ \rho \displaystyle\int_V  \left[  \nabla \cdot \dfrac{(\hat{r})}{r^2} \right] dV  =0$$ Now we do it in another way by applying Gauss divergence theorem and solid angle formula: $\nabla \cdot \vec{E} =\nabla \cdot     \left[    k\ \rho \displaystyle\int_V  \dfrac{dV}{r^2}  (\hat{r})  \right] =k\ \rho \displaystyle\int_V \left[ \nabla.  \left( \dfrac{\hat{r}}{r^2}   \right)  \right] dV =k\ \rho \displaystyle \iint_S \dfrac{(\hat{r})}{r^2} \cdot \vec{dS} =4\ \pi\ k\ \rho$ Why am I getting different results?
120563
I am trying to formally learn electrodynamics on my own (I only took an introductory course). I have come across the differential form of Gauss's Law.   $$ \nabla \cdot \mathbf E = \frac {\rho}{\epsilon_0}.$$  That's fine and all, but I run into what I believe to be a conceptual misunderstanding when evaluating this for a point charge.   I know the math looks better in spherical coordinates, but I will be using Cartesian.   So when I calculate the divergence I obtain:  $$ \nabla \cdot \mathbf E = \nabla \cdot kQ\langle\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}},\frac{y}{(x^2+y^2+z^2)^{\frac{3}{2}}},\frac{z}{(x^2+y^2+z^2)^{\frac{3}{2}}}\rangle = \frac{-3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}+\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}}.$$  This can further be simplified:  $$\frac{-3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}+\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}} = \frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}}-\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}} = \frac{3-3}{(x^2+y^2+z^2)^{\frac{3}{2}}}.$$   Now instinctively I would say that 3-3 is zero and then the while thing is zero everywhere. I am confused as to why (purely mathematically) this expression is not equal to zero at the origin. I completely understand why it physically has to be that way. And I also understand that it is modeled with the delta dirac function. But what (again, mathematically) is stopping me from saying that equation is just zero even at the origin?
7911
I am trying to understand the formula \begin{equation} \nabla^2\left(\frac{1}{|{\bf r}-{\bf r}'|}\right) = - 4 \pi \delta(\bf{r}-\bf{r}'), \qquad\qquad {\rm (I)} \end{equation} where ${\bf r}=(x,y,z)$. This is something heavily used in electrostatics and the steps to 'show' this is often the following: The first derivative reads \begin{equation} \nabla \frac{1}{| {\bf r} - {\bf r}' |} = - \frac{ {\bf r} - {\bf r}'}{| {\bf r} - {\bf r}'|^3} \end{equation} And taking the second derivative gives zero, except for the singularity at ${\bf r} = {\bf r'}$. Then from the divergence theorem we have \begin{equation} \int dV \,  \nabla^2 \frac{1}{| {\bf r} - {\bf r'}|} = \int dS  \,\,{\bf n} \cdot ( \nabla\frac{1}{|{\bf r} - {\bf r}'|}) = -4 \pi \end{equation} where the integration is performed over a sphere centered at ${\bf r}'$. Q1: Is there a more direct proof for equation (I)? Then my main question is about the separate second-order differentials. For instance, we can obtain, by direct computation \begin{equation} \partial_x^2 \, \frac{1}{| {\bf r} - {\bf r}'|} = \frac{ 3 (x-x')^2 }{| {\bf r} - {\bf r'} |^5} - \frac{1}{| {\bf r} - {\bf r}'|^3} \end{equation} Q2: Should there be a $\delta$ function on the r.h.s of this equation?
81697
If you have a conducting hollow sphere with a uniform charge on its surface, then will the electric field at every point inside the shell be 0. The reason the electric field is 0 at the center is clear from the symmetry of the sphere, but for a point at a certain distance from the center, shouldn't a net electric field exist?  I saw a similar question on the site but it was vaguely explained.
78037
I have, for instance, a problem with a spherically symmetric charge distribution. I deduce here, in order to solve the problem easily, that the corresponding electric field must be symmetric. How is this type of deduction rigorously justified?
36150
If we write out Maxwell's equations with magnetic charges, we get $$ \begin{align} \nabla \cdot \mathbf{E} &amp;= 4 \pi \rho_e \tag{1}\\ \nabla \cdot \mathbf{B} &amp;= 4 \pi \rho_m \tag{2}\\ -\nabla \times \mathbf{E} &amp;= \frac{\partial \mathbf{B}}{\partial t} + 4 \pi \mathbf{J}_m \tag{3}\label{Eq:Faraday}\\ \nabla \times \mathbf{B} &amp;= \frac{\partial \mathbf{E}}{\partial t} + 4 \pi \mathbf{J}_e \tag{4}\label{Eq:Ampere} \end{align} $$ In particular, Faraday's law \eqref{Eq:Faraday} contains a minus sign that Ampere's law \eqref{Eq:Ampere} does not. This always struck me as odd because it's often said the fields are dual to each other (i.e. you can replace E with B and &quot;get the same result&quot;), but that requires a bit of mental recalibration to accommodate that minus. So I'm curious what the origin of that negative is and what it means. Are there any intuitive explanations for how to think about it?
22203
Consider the vector field $$\vec{\mathrm{F}} = \frac{\hat{\mathrm{r}}}{r^{2}},$$ then the divergence of this field is: $$\vec{\nabla}\cdot\left(\frac{\hat{\mathrm{r}}}{r^{2}}\right) = 4\pi\delta^{3}(\mathrm{\vec{r}})$$  What is the proof of this relation?
45750
is an amazing law. It is extremely useful (as far as problems framed for it are concerned :D. I do not have a real world-problem solving experience of using Gauss' Law).  However, I don't really understand why it works. What my doubt is, is this:     In the God-equation:      $$ \oint_s\vec{E}\cdot d\vec{s} = \dfrac{q_{enc}}{\epsilon_\circ} $$      $\vec{E}$ is the field at the area element $d\vec{s}$, and that field   is due to all the charges in the configuration. But the $q_{enc}$ is   the algebraic sum of only those charges that are enclosed within the   Gaussian surface. That is, to the net flux, only those charges affect that are within the surface. The contribution of the outside charges somehow cancels out. Why does it work that way?   If I knew a lot of advanced math and vector spaces and such fancy stuff, I wouldn't have asked this question. I am looking for an intuitive way of understanding its proof - if it is possible to understand without that depth of math. I do know some basic integration and surface integrals and all that (otherwise I wouldn't be able to use Gauss's Law, obviously!). So, if your answer involves these things, I'd be fine. But please don't bring in the divergence theorem etc. I tried to decipher the math behind it on wikipedia, and didn't understand a thing.
84745
The electric force (Coulomb's law) on a point charge $Q_2$ due to $Q_2$: \begin{gather*} \mathbf{F}_{12}=\frac{Q_1Q_2}{4\pi\epsilon_0}\frac{\mathbf{r}_2-\mathbf{r}_1}{|\mathbf{r}_2-\mathbf{r}_1|^3} \end{gather*}  The seperation vector between the charges is $\mathbf{r}_{12}=\mathbf{r}_2-\mathbf{r}_1=(x_2-x_1)\hat{\textbf{i}}+(y_2-y_1)\hat{\textbf{j}}=(z_2-z_1)\hat{\textbf{k}} $.  Does this mean that the force is a vector field or just a vector?   Is the force a function of $\mathbf{r}_{12}$, so $\mathbf{F}_{12}=\mathbf{F}_{12}(\mathbf{r}_{12})$?
105537
My question may be very basic, but I can't think of a reasonable explanation for this.  Consider a solid charged sphere. Now, we have an electric field inside the solid sphere, but at any particular point there are charges infinitesimally close to the point. As I see it, these charges should contribute to an infinite value of the electric field.  One might possibly argue that at any given point, infinitely-close charges are distributed uniformly everywhere and hence their effects cancel out. But at the surface of the sphere the uniformity of infinitely-close charges is not present and hence here we must have infinite field, but that is not the case as well! Where are we going wrong?  Also, if we think of electric field at that point in continuous charge distribution, the contribution of electric field might exist finitely from nearby charges with both charge and distance between them tending to zero, but for the very small charge present exactly there, the electric field contribution from itself must be infinite with charge being finitely small and distance being exactly 0. What's wrong here?  I have strong suspicions, is it sensible and correct to define electric fields at such a close vicinity of other continuous charges?
101527
I was teaching kids about how to find electric field using the superposition principle for continuous charge distributions. I thought maybe I should derive the formula for electric field due to a finite rectangular sheet of charge of charge on the surface $S$, where $$ S = \left\{(x,y,z)\in \mathbb{R}^3 \mid -a/2&lt; x &lt; +a/2; -b/2&lt; y &lt; +b/2 ; z = 0 \right\} .$$ However, I got stuck at the following integration. $$ E(0,0,r) = \frac{\sigma r}{4\pi\epsilon_o}  \int_{x=-a/2}^{x=+a/2}\int_{y=-b/2}^{y=+b/2} \frac{dx dy}{(x^2+y^2+r^2)^{3/2}}, $$ where $\sigma$ is the surface charge density. Note: This integration can be done if $a$ or $b$ or both are very large i.e. $\infty$ in which case we get usual result of $E=\frac{\sigma}{2\epsilon_o}$ So my question is, Can this integral be calculated? If not then what method would I use to find the electric field in this case. Also It would be greate if anyone can comment on how to find the electric field by directly solving the poisson equation.  Consequently if we take case of finite disk the following is the resulting integration. $$ E = \frac{\sigma r}{2\epsilon_o}  \int_{\xi=0}^{\xi=R} \frac{\xi d\xi}{(\xi^2+r^2)^{3/2}} $$ which can be solved as $$ E = \frac{\sigma}{2\epsilon_o} \left(1- \frac{r}{\sqrt{r^2+R^2}}\right) $$ Now by taking the limit $R \rightarrow \infty$ we can show that $E \rightarrow \frac{\sigma}{2\epsilon_o}$.
5782
Calculate: $$\int_{0}^{2\pi}e^{R{ {\cos t}}}\cos(R\sin t+3t)\mathrm{d}t$$ My try: $\displaystyle\int_{0}^{2\pi}e^{R{ {\cos t}}}\cos(R\sin t+3t)dt\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\cos(\mathfrak{I\textrm{z}}+3(-i\log z)dz\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}}^{(\mathfrak{I\textrm{z}}+3(-i\log z))i}dz\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}^{\mathfrak{I\textrm{z}}}}z^{3}dz\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}^{\mathfrak{I\textrm{z}}}R}z^{3}dz$ and there is nothing here that is not holomorphic, therefore according to Cauchy theorem it must be exactly $0$.
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?
44826
A positive and negative charge are some distance d away from each other (Electric dipole configuration) Consider the electric field lines in between them. That is, there is a straight line electric field vector that points from (+) to (-) along the d direction. There is also an arbitrary electric field line exiting the (+) charge by some angle alpha with respect to the d direction and it enters the (-) charge by some angle beta with respect to the d direction. Assume alpha does not equal beta. What are the angles?  This was a question posed by my professor to think about. At first I thought the angles would be equal so I didn't really understand his question. I think the case that they would be equal would be when the positive charge magnitude is equal to the negative charge magnitude (I may be wrong). He did not tell us what the magnitude of the (+) and (-) charges so I think this may be the main point. Would there be a way to calculate the angle? Maybe a ratio? This is mainly for my own curiosity any insight would be appreciated!
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.
29023
The curl in cylindrical coordinates is defined:  $$\nabla \times \vec{A}=\left({\frac {1}{\rho }}{\frac {\partial A_{z}}{\partial \varphi }}-{\frac {\partial A_{\varphi }}{\partial z}}\right){\hat {\boldsymbol {\rho }}}+\left({\frac {\partial A_{\rho }}{\partial z}}-{\frac {\partial A_{z}}{\partial \rho }}\right){\hat {\boldsymbol {\varphi }}}{}+{\frac {1}{\rho }}\left({\frac {\partial \left(\rho A_{\varphi }\right)}{\partial \rho }}-{\frac {\partial A_{\rho }}{\partial \varphi }}\right){\hat {\mathbf {z} }}$$  For vector fields of the form $\vec{A}=\frac{k }{\rho}\hat{\varphi}$  (plotted below), $A_z=A_\rho=0$ and $A_\varphi = k\rho^{-1}$, so the resulting field has zero curl. But choosing $k=\frac{\mu_o I}{2\pi}$ results in the correct solution for the magnetic field around a wire:  $$\vec{B}=\frac{\mu_o I}{2\pi R}\hat{\varphi}$$  This field cannot be curl-free because of Maxwell's equations, Ampere's law, etc. So I must have made a mistake somewhere: Why am I calculating this field to be curl-free?
119651
Recently I learned that electrostatic potential energy of a system of charges can be calculated like so:  $$E = \int \frac{1}{2} \epsilon_0 \mathbf E^2 dV.$$  However, for a system of two point charges Q1 and Q2, we have  $$E = \int \frac{1}{2} \epsilon_0 (\mathbf E_1 + \mathbf E_2)^2 dV = \int_{}^{}\frac{1}{2}\epsilon_0 \mathbf E_1^2 dV + \int_{}^{}\frac{1}{2}\epsilon_0 \mathbf E_2^2 dV + \int_{}^{}\frac{1}{2}\epsilon_0 * 2 \mathbf E_1 \cdot \mathbf E_2 dV$$  from which we subtract the first two terms because they are already properties of the charges themselves.  Question:  Why do we not need to do something like this for charge distributions? Aren't charge distributions made up of many many little charges of magnitude $e$? Even if we treat the charge distribution as perfect continuous, won't there be infinitely infinitesimal charges?
106611
Amperes Law: $\oint\vec{H}.d\vec{l} = I_{enc}$  Is the current, $I_{enc}$, equal to 0 outside of the conductor ($\rho &gt; a$)  because the current density, $\vec{J}$, outside is 0?