Uniform fields are created by setting up a potential difference between two conducting plates placed at distance (d) from one another:   The electric field equation is:   $\vec{E}=kQ/r^2$  so if you have 2 plates(+,-) with distance d. then the electric field should be:  $\vec{E_{tot}}=\vec{E_{+}}+\vec{E_{-}}$  $\vec{E_{+}}= k(Q)/(d)^2$  $\vec{E_{-}}= k(-Q)/(1-d)^2$  $\vec{E_{tot}}=k(Q)/(d)^2 + k(-Q)/(1-d)^2 $  but isn't this field proportional to:   $\frac{1}{d^2} +\frac{1}{1-d^2} $  which is not a constant?  I imagine the filed to be stronger near the plates and weakest in the middle.
36038
I know from Gauss law, it is $\vec{E}=\dfrac{\sigma}{2 \epsilon_0}(\hat{n})$ at all points. But it doesn't make sense because of the inverse square nature of electric field which suggests if you move further away from the plane, electric field must reduce.   Then why is electric field of an infinite plate constant at all points?
87469
As I remembered, at the 2 poles of a battery, positive or negative electric charges are gathered. So there'll be electric field existing inside the battery. This filed is neutralized by the chemical power of the battery so the electric charges will stay at the poles.  Since there are electric charges at both poles, there must also be electric fields outside the battery. What happens when we connect a metal wire between the 2 poles of a battery? I vaguely remembered that the wire has the ability to constrain and reshape the electric field and keep it within the wire, maybe like an electric field tube. But is that true?
104848
So here are the two of Maxwell's laws that I am interested in:  So we have the simple circuit (from google):   So, before the system goes into steady-state we know that charge slowly accumulates on the plates of the conductor. So the charge on the plates gets bigger and bigger while the charges that carry the current get smaller and smaller, so the current gets weaker. Applying Ampere's law on the wire we find the induced magnetic field due to the current $I(t)$ that penetrates the surface $\Sigma$ (see the integral of $\mathbf{J}\mathrm{d}\mathbf{S}$) and not due to an electric field. Now, this induced magnetic field is changing with respect to time (because current is changing). But from the Maxwell-Faraday equation we conclude that this changing magnetic field will produce an electric field which again changes with respect to time. And then we have another induced magnetic field due to that changing electric field. And the cycle goes on. So, am I right? And if I am, when does this stop? And how does it changes the way I calculate each induced field? Does it have to do with electromagnetic waves?
30268
$$\mathbf D = \varepsilon \mathbf E$$ I don't understand the difference between $\mathbf D$ and $\mathbf E$.  When I have a plate capacitor, a different medium inside will change $\mathbf D$, right? $\mathbf E$ is only dependent from the charges right?
29129
Since the electric field inside a conductor is zero that means the potential is constant inside a conductor, which means the "inside" of a conductor is an equal potential region.   Why do books also conclude, that the surface is at the same potential as well?
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 $\mathbf{F}_{12}=\mathbf{F}_{12}(\mathbf{r}_{12})$?
33260
When I look at electric or magnetic fields, each of them has a constant that defines how a field affects or is affected by a medium. For example, electric fields in vacuum have a permittivity constant $ϵ_0$ embedded in the electric field expression of a point charge: $E = q/4π ϵ_0r^2$. However, if I put this point charge in some dielectric that has a different  permittivity constant $ϵ$, the value of the electric field changes. On a similar note, magnetic fields behave very similar but have the permeability constant $μ_0$ instead.   From my understanding, I believe that this is not the case for gravitational fields since the universal gravitational constant $G$ is consider to be a fundamental constant. So I am assuming that even though gravitational fields do operate in different types of mediums, this somehow doesn’t affect the gravitational field value. My question is why is this the case, that is, why isn’t there a permittivity-type constant for gravitation?
87857
I've learn some electronic knowledge at school and so confused by some questions.  Q1. IS it possible a flow of electricity without an electric field?  Q2. I've learn that moving of particle with electric charge in electric field makes potential difference. Then, let's suppose no resistance in wire. IF Q1 is true, wire with a flow of electricity makes an electric field, and in the electric field, an electron moves. Then potential difference must exist. But when measured, there is no potential difference. ( I learn there is ESR in battery, but I wonder whether moving of electron dosen't make any potential difference or not.)  Sorry for my bad English, but I hope someone could help me to find answer.
47102
My textbook says that when a conductor is placed in an electric field, the electrons in it realign so that the net electric field inside the conductor is zero. There isn't a proof for this. It merely states that the electrons drift opposite to the applied electric field and create an opposite electric field inside the conductor canceling the external one.   I imagine Gauss's Law would provide a simple proof. If we try enclose a point inside the conductor in a gaussian surface, and note that the enclosed charge is zero, so the flux through the surface is zero and so is the electric field at the surface.   When I looked back at my reasoning, I realized that I had assumed that the charges induced on the conductor always stay at the surface, which I had no way of proving. How do I prove this formally?
119763
Say we have a hollow conducting sphere (with some finite thickness).  If this object has an excess charge amounting to +Q coulomb, and there is no extra electric field in the surroundings (due to other charges), how will the charge be distributed?  Intuitively it seems that the charge will be symmetrically distributed..(all other cases seem too ugly), but in this case how is charge inside the conducting surface zero? Does this have something to do with solid angle and the fact that two cones with vertex at a point in the conductor will have a special relation of the (charge/distance^2) factors? (more charge is distance is more...)  If this is the case can someone please prove that the effect of the two corresponding cones will cancel each other out?
87612
I was reading an article about resistivity when I faced this equation:  $$V_P-V_O=\int_P^O E.dr$$  The equation is in one dimension and $P$ and $O$ are points on the $x$ axis.  The Electric field is oriented in the $+x$ direction ans has the formula $c\frac1r$ where $c$ is a coefficient and $r$ is the distance form the origin of the coordinate system.  The vector $\vec r_P - \vec r_O$ has the same direction as that of the electric field.    Now, Why do we put $P$ below the integral and $O$ above it? Can someone explain the equation for me?
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?
114535
Electric flux is a defined quantity that is proportional to the no. of field lines passing through a given area element for a given electric field. It is not proportional to the relative density of field lines, which would supply information regarding the strength of the field at that point. Electric flux, it seems to me, does not supply us with any practical information. It seems to me that electric flux is a quantity defined and modeled specifically for Gauss' law, to introduce some kind of mathematical elegance to it and to introduce an additional visual aspect to the concept of electric fields. Perhaps this is why for symmetrical situations especially, Gauss' Law can be used to easily determine the electric field due to the given charge distribution. Am I wrong here? Is there a physical significance to electric flux that I do not understand?
8080
When we're dealing with problems in electrostatics (especially when we use Gauss' law) we often refer to the density of electric , which is inversely proportional to the radius in the case of a single point charge (all field lines are directed radially).   My question may sound dumb, despite the fact that this concept is quite intuitive, but if you think about it there is actually an infinite amount of field lines which we can draw everywhere (one through every point), so speaking of an area where field lines are more "dense" or "sparse" doesn't make much sense to me.  With this in mind, why can we still use such a concept? Why does it really work?
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.
43122
I know gradient shows you the direction in which function is increasing most. So does electric field direction show us in which direction potential energy function is decreasing most....  My efforts : we know electric field lines go from positive to negative charge. So that's where value of electric potential energy function is decreasing. As we know electric field lines show us the electric field direction. So electric field points out the direction in which potential energy function is decreasing so it's the negative gradient vector of the potential energy function. Am I right with this reasoning?
18687
1. How is the electric field set up in a wire by a battery so that we can  assume that potential at a node is constant?  2. What happens to the electric field across a resistance which makes voltage 'drop'?     Assume the following circuit:      The battery sets up some electric field throughout the wire, with a component along (parallel to) the wire (otherwise, there is no current). Zooming into the wire,   Here I display only the component of the electric field and electron velocity that is along the wire. As the electron moves a distance d, the work done on it by the electric field is Eqd. This work is non-zero. In other words, electric potential energy of the electron varies with the distance it travels along the wire - and thus so does potential. How, then, is potential on a node the same?    Now, add a resistive load:    The potential at all points on the wire above the battery is 10V, below it - 0V. Considering the above - how is this so?    NOTE: I am aware that a non-ideal wire has resistance. My question lies with what resistance has to do with a difference in potentials in the first place - and why voltage drop is not CAUSED BY a charge's distance from the terminals of the battery.    Exact duplicate it is.