A very common question about Gauss law is being asked in schools or Internet:      Why the electric field of an infinite plane is independent on distance?     Actually I was one of who asked this question and I found an answer,  and I will explain it in the next paragraph.  A very important thing we have to remember when think about this kind of problems is that the electric field is a vector, and the $Ex$ vector only effect on it because $Ey$ is cancelled by another charge.  Since we go further. the electric field is being weaker, also the angle $θ$ is being less too So $\cos θ$ is getting larger so that compensated the lack of electric field.   Now, why we cant apply this argument on an infinite line, as you see in the picture?    Please don't explain it using Gauss law because it's hard to visualize it, because the number of imaginary concepts like : (Flux,  Electric field,  epsilon ... etc.) So it's hard to realize.  Finally,  please don't mark my question as a duplicate because I didn't understand Eeko explanation.
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I understand that with an infinite plane, as you get closer, the infinitesimal contributions to the electric field become greater in module. The direction of the vectors become less perpendicular to the plane as you get closer, hence reducing the overall electric field in the perpendicular direction. Both effects offset each other such that the electric field is the same at whatever distance.  However, wouldn't this same argument apply to an infinite line? Gauss's law shows that the electric field of an infinite line depends on the distance, but supposedly the same would happen as with the infinite plane, and yet it doesn't.  Does anyone have an explanation for this?
2949
My professor said that the electric field is zero wherever two equipotential surfaces intersect. I can't come up with a reason why.  He also claimed two equipotential surfaces cannot intersect as that would give two different potentials at the same point. Why can't there just be two different equipotential surfaces with the same potential that intersect or touch?
465
What's the difference between magnetic field and electric field lines?  Does $d\vec B$ point in the direction you would experience a force if you were a moving charged particle at that point? I know for a fact the electric field and electrostatic force are parallel, but since $F = q\ \vec v \times \vec B$ for magnetic forces. Does that actually mean though that if, say, in the following graphic:    Where the dot is a wire going in and out of the page, that the $d\vec B$ vectors are not showing how a moving charged particle will move in the field?   Basically, I feel justified in thinking electric field lines communicate how a test charge will move at any given point in space.  How can I juxtapose this with magnetic field lines>
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My question and context with explanation are given below. Thank you in advance.   Faraday’s Law () states that a time-changing magnetic field vector induces a curl of the Electric field around that point. However it does not specify how quickly the necessary spatial gradient is developed in the Electric Field to satisfy that curl. Of course, the Electric Field vectors around that point cannot just jump to the required values instantaneously in accompaniment as soon as the magnetic field starts changing at a fixed rate, since it is required that the Electric Field has a sensible rate of change to in turn induce a curl of the magnetic Field and continue the propagation of the wave. It would seem the exact rates at which the Electric and Magnetic Fields are changing is critical in the mechanism of an EM wave’s propagation. How would I then calculate the rate of change of E somewhere along the wave given I know the rate of change of the B vector that began it, and given the wave is restricted to one dimension for simplicity? There should after all be some delay in how the EM wave moves from point to point in space, seeing as it moves at the finite speed of causality (3*10^8 m/s), as shown in the first image:    Some of my ideas so far have been that perhaps the rate of change of E is determined such that it exactly cancels the growth of the time-changing E vector that gave rise to a curl of B before, which developed at such a rate to induce a curl of E, which demands this rate of change of E which I am talking about now. In this way the wave might level out the growth of the fields in its wake and spread the change further outwards, in the same way smoothening out a crease spreads it elsewhere. However I am stuck on how to show this rigorously.  I am open to any difficulty of maths, so long as the answer provides the explanation. Below are some diagrams to help visualize what I am talking about:    Faraday's Law requires that a curl of E is developed, which under these restrictions requires a spatial gradient of E to appear across that point. Since the field across that point was previously in some other state, it is required that the Electric Field vectors that are infinitesimally close to that point undergo a rate of change through time to achieve the desired state of curl. The question is, how quickly is this desired state of curl achieved?
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Electric field of a point charge drops off inverse squarely but that of a dipole varies as $1/r^3$. Why the field drops off more quickly for a dipole?
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This is a question from the second chapter (Problem 2.55) of Griffiths 'Introduction to Electrodynamics' fourth edition. The question says that if in a region, the electric field $E$ exists such that, $E_x = ax$ where $a$ is a constant, $E_y=0$ and $E_z=0$, what is the charge density. The charge density is given by the differential Gauss law, where, $\nabla.E = \dfrac{\rho}{\epsilon_0}$. This gives $\rho=\epsilon_0a$. Since $a$ is a constant, the charge density is uniform everywhere in the region. Now Griffiths further asks how could this uniform charge density produce an electric field that is parallel to just $X$ axis?  My intuitive answer is if there is an infinite slab like uniform distribution of charge where the slab has infinite length and breadth but a finite thickness, if the thickness is along the $X$ axis, the electric field will be along $X$ axis given as $E_x = ax$, $E_y=0$ and $E_z=0$.  However Griffiths himself answers this question in the instructors solution manual as given below.  While I also agree to Griffiths on a broader picture, I want to know if my intuitive solution of an infinite slab of finite thickness wrong?
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I read this in my textbook: A charged particle or object is not affected by its own electric field.  Since I find this completely unintuitive and my mind is yelling "wrong! wrong! how could a particle even distinguish between its own field and the external fields?" I would really like to hear an explanation about why this is true, if it is.  (A corollary question that comes to my mind after thinking about this is... What about gravity and mass considering the analogous situation? And the other two forces?)
22962
We have studied so far that electric field inside a conductor if no charge is placed inside is zero. But we know that every conductor has only a limited number of electrons.  What happens when ALL the electrons have aligned to cancel the field inside conductor under application of external field and then we increase the field a little bit more certainly there are no more electrons to cancel the field and this additional field must be present inside the conductor now.  I know that practically under electric fields of magnitude that can drag out all electrons from the conductor's body to surface field emission and/or electrical breakdown would be taking place, But is this phenomenon theoretically possible ?
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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?
18203
I would like to clear things up: How exactly the electric field inside a conductor is zero? Let  a really "powerful" electric field be outside of it, how can the "few" charges in a conductor balance the field? From my point of view, it's the conductor surface which stops their movement. I'm obviously missing something fundamental here. Thanks in advance.
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Let's say we have a cube of side length $a$ as a Gaussian surface in an electric field (in vacuum) and then we raise the strength of the electric field. As I understand, this rise in the field strength would travel through the field at the speed of light. This means that for a duration of $\frac{a}{c}$, the field strength, and thus the flux, would differ at the 2 opposing faces of the cube. This implies that the flux through the surface is non-zero and thus the charge inside it must also be non-zero by Gauss' Law. How is this possible?
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My physics textbook says "Electrostatic field lines do not form closed loops. This is a consequence of the conservative nature of electric field." But I can't quite understand this. Can anyone elaborate?
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Say we have 2 charges set up like above. Why is $V_3=0$ ?   I can understand mathematically that the scalars fields cancel out, but I don't understand it physically. If I place a positive charge in between the 2 charges, it has electric potential energy because it is feels a force towards the negative charge and repelled by the positive and gains kinetic energy. However because the Electric Potential is 0, thus the electrical potential energy is 0.  How can $V_3 =0V$ if a charge placed in between the two charges will gain kinetic energy?
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In MIT's 8.02 course, it is shown in lecture 3 that we can derive Gauss's Law from Coulomb's to get  $ \phi = \oint \vec{E} \cdot \vec{dA} = \frac{Q_{enc}}{\epsilon_{0}} $  However, in the lecture, it was assumed that there were no charges outside the surface. Later, Gauss's Law was used and the charges outside the sphere were ignored.  I've been thinking about it and came across .  The first answer by user levitopher helped a bit. I think that the lecture's argument still holds when there are charges outside because $ \vec{E} $ is added vectorially.  Suppose we have $ Q_{enc} $ in a sphere and $ Q_{out} $ outside the sphere. Then,  By definition, $ \phi = \oint (\vec{E_{Q_{enc}}} + \vec{E_{Q_{out}}}) \cdot \vec{dA} $ $ = \oint \vec{E_{Q_{enc}}} \cdot \vec{dA} + \oint \vec{E_{Q_{out}}} \cdot \vec{dA}$ $ = \oint \vec{E_{Q_{enc}}} \cdot \vec{dA} + 0 $ $ = \frac{Q_{enc}}{\epsilon_{0}} $  This assumes that $ \oint \vec{E_{Q_{out}}} \cdot \vec{dA} = 0 $ because any field lines that go in must come out (hand wavy but I'm just going to accept it for now).  I'd just like to know if this argument is sound.
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I did lot of searching but couldn't find any textbook nor any webpage which could clarify my doubt. Maybe my doubt is insanely stupid and I was dumb for not realizing it at the first place.  The negative of the work done by the electrostatic field in bringing a charge from infinity to a point is called electric potential.   Let us assume that there is a positive charge at the origin. Let work done by the external force to bring a positive from infinity to a point P close to the origin be W. And hence work done by the field to bring a positive charge from infinity to the point P will be -W.  If I did work W to bring the charge from infinity or if the field brought the charge from infinity, either way the change in potential energy will be same.  In case of work done by the field, we say that the work done is stored in the form of electrostatic potential energy.  In case of work done by the external force, we say that the work done was positive and the energy was taken by some external source.  In the above case, I did work W in bringing it &amp; the field did work -W on the charge trying to push it away so I did positive work of W &amp; the field did negative work of W. We can say that the work done by the field was stored as potential energy.   Where did the work done by the external force (me) go?  I can summarize the whole doubt in the following line: I did work W to bring an charge towards another unlike charge, and therefore the field also did work -W, the net work done on the charge is 0. But there is still a change in potential energy. Why did the potential energy change? Isn't conservation of energy being violated here?  Please clarify my doubt (I do understand that there is a horrible conceptual error in one of my arguments but I do not know which one it is). If I am not wrong this has nothing to do with electrostatics rather has to deal with field theory/inverse square law. I believe I will be encountering the same problem again when I would be studying another force, maybe gravitational, which obeys inverse square law.  Thank You
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I am a bit lost how one has deduced the formula for electric field with electric dipole because of some inconsistency between different sources. The  contains a delta function in the formula, but  gives only the formula without the delta function,  $$\bar{E} = -\nabla \phi = \frac{-p}{4\pi \epsilon_{0} r^{3}} ( \hat{e}_{z}-3\hat{u}_{r}(\hat{u}_{r} \cdot \hat{u}_{z}) ) $$  Is it intentionally left out?  My notes mention that the dipole is with a circular thing $P=Q d$ where $d$ is distance and $Q$ is charge.
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Suppose you place a point charge $+Q$ at the corner of an imaginary cube.  Since electric field lines are radial, there is no flux through the three adjacent (adjacent to the charge) sides of the cube. However there is some amount of flux passing through the other three sides of the cube (flowing out of the cube).  We can estimate that the flux through these three surfaces combined is equal to $Q/(8\epsilon)$. As, if you consider $7$ other cubes having the charge at the corner, each of them would have equal flux flowing out by symmetry and since the total flux through the $8$ cubes is $Q/\epsilon$, each cube would have a flux of $Q/(8\epsilon)$.  Now apply Gauss' law to the cube, and we find that the cube encloses a charge of $Q/8$.  This means that 1/8th of the charge belongs to this cube.   But the charge we placed was a point charge with no dimensions. It cannot be split into parts.  What is wrong?