This kind of a configuration struck me when I read  question.   The original question asked how the force on the two charges would change if a metal plate was inserted between them. This question addresses what would happen if it was an infinite plane (those are so much more convenient).  I gave it some thought and the result is rather interesting.   EDIT: This question is very different from the original in that:   The infinite plane is grounded, a plate with definite dimensions does not have to be One can't use image charges with the plate If the plate is uncharged, positive charges will accumulate on the edges and create their own field, in case of an infinite plane these positive charges are at infinity and do not affect the configuration. A plate does not provide electrostatic shielding or a simple boundary region to apply the uniqueness theorem in. The charge distribution on a finite plate is much harder to model.
123277
If a thin metal plate is placed between two charges $+q$ and $+q$, will this cause a change in the electrostatic force acting on one charge due to another? What is the concept behind this? What will happen if the metal plate is thick?
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!
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?
48633
In electromagnetism books, such as Griffiths or the like, when they talk about the properties of conductors in case of electrostatics they say that the electric field inside a conductor is zero.  I have 2 questions:    We know that conductors (metallic) have free electrons which randomly moves in all directions, so how come we can talk about electrostatics which by definition means stationary charges? When the textbooks try to show why the electric field inside a conductor is zero they say let us put our conductor in an electric field. What happens then is that there will be an induced surface charge density which consequently induces an electric field within the conductor such that the total electric field within the conductor will be zero. That is perfectly understood, but my problem is the following: the original claim was that the electric field within a conductor is 0, not the electric field after putting the conductor in an external electric field it became zero. I do not understand the logic!
86821
As I understand, the different voltage between two object is caused by the total different charged between two object. Thus, the higher extra electron between two opposite charged object, the higher voltage between them, isn't it? So if i have two battery or capacitor connect in parallel, the negative connect to negative, the positive connect to the positive. Caused the charge from each plate to be double. So that's as my expected, it must be double voltage, but ironically why it's out of my expected and the reality is vice-versa?
25871
I know that the $E$ field inside a conductor is zero. What happens if I put a source of charge inside the conductor?  Say the conductor was spherical centered on the origin and there exists a charge at the origin.  Then by Gauss's Law, there exists a non zero field inside the conductor.  Or is it the case that even though this charge is within the sphere, it will redistribute evenly over the surface and hence the E field inside is zero?
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?
16061
I am confused as to whether/how capacitance changes when each plate has a different charge.  For example, consider a coaxial cable and put $20Q$ on the outer cable, and $-Q$ on the inner. Or how about concentric spheres, grounding either the inner or the outer?
82362
Actually my question comes from Gauss law which says charges outside a closed surface, doesn't effect on the whole electric flux of the surface.  The reason of that is that the solid angle of a closed surface while measuring it from outside of the surface, is zero. ( on the other hand, the solid angle of a closed surface from inside the closed surface is $4\pi$)  My question is why solid angle of a closed surface from exterior  is zero? Please don't use Stokes' theorem!
81840
A metal rod of length $l$ is placed (as shown in figure) in a cylindrical time varying magnetic field. Find the potential difference across it.   My answer:  As electric fields as forming circles around the centre with magnitude $$\frac12r\frac{dB}{dt}$$ whose magnitude is same along any chord at a fix distance from centre: $$\Delta V=E.l=\frac12r\frac32at^{\frac12}.l$$ $$\Delta V=\frac34arlt^{\frac12}$$  My question: Is $\Delta V =V_M-V_N$ or $\Delta V=V_N-V_M$ $\bf OR$ equivalently which side is at high potential?  Possible    $(i)$ As Field lines are circular from $M$ to $N$, $M$ must be high potential. $(ii)$ As this field line apply a force on electrons- in the rod- towards left, then $M$ must have excess $e^-$ and $N$ must have deficiency; so $N$ would be high potential as $M$ is $(+)$ and $N$ is $(-)$.
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.
34990
Would the field in a cavity inside an arbitrary conductor with some charge, which is NOT SYMMETRIC be zero?   My book only says that field inside a cavity is always zero when there is no charge in cavity. But they prove this by taking a sphere with a symmetrical cavity and using Gauss law. But how can I prove this in cases where I can't take $E$ (field) out of:   $$\oint\pmb{E}\cdot d\pmb{A}=\frac{Q_{enclosed}}{\varepsilon_0}.$$
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?
75650
For a localized charge distribution the potential is set to zero far away from the charge distribution (at infinity)  Now, when grounding a conductor, i.e. connecting it to Earth, it is said that we are setting its potential to zero.   Why Earth's potential is zero? And if it is zero, is it zero even if the potential is still zero at infinity?
113104
When we find the electric field between the plates of a parallel plate capacitor we assume that the electric field from both plates is $${\bf E}=\frac{\sigma}{2\epsilon_0}\hat{n.}$$ The factor of two in the denominator comes from the fact that there is a surface charge density on both sides of the (very thin) plates. This result can be obtained easily for each plate. Therefore when we put them together the net field between the plates is $${\bf E}=\frac{\sigma}{\epsilon_0}\hat{n}$$ and zero everywhere else. Here, $\sigma$ is the surface charge density on a single side of the plate, or $Q/2A$, since half the charge will be on each side.   But in a real capacitor the plates are conducting, and the surface charge density will change on each plate when the other plate is brought closer to it. That is, in the limit that the two plates get brought closer together, all of the charge of each plate must be on a single side. If we let $d$ denote the distance between the plates, then we must have $$\lim_{d \rightarrow 0}{\bf E}=\frac{2\sigma}{\epsilon_0}\hat{n}$$ which disagrees with the above equation. Where is the mistake in this reasoning?  Or more likely, do our textbook authors commonly assume that we are in this limit, and that this is why the conductor behaves like a perfectly thin charged sheet?
85178
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?
33004
It is always told as a fact without explaining the reason. Why do two objects get charged by rubbing? Why one object get negative charge and other get positive ?