New answers tagged

2

Looks like, in general, $\frac{\partial E_x}{\partial y}\neq\frac{\partial E_y}{\partial x}$, so $\overrightarrow{E}$ cannot be a gradient of any decent function. Therefore, I don't think the problem has an unambiguous answer.


-2

V=E/L. V=Voltage. E=Electric field magnitude. L=Length between the points A and B.


0

If the electric field is not dynamic (if it does not depend on time) there is no magnetic term and the electrical potential $\phi$ is defined such that: $E = - \nabla \phi$. You should therefore integrate the electric field $E$ on the given domain (I would recommend using a software).


0

Suppose the surface charge densities on the bottom plate is $\sigma$ and on the top plate $-\sigma$, then the electric field due to the bottom plate is $\frac{\sigma}{2 \epsilon_0}{\bf n}$ and that due to the top plate $-\frac{\sigma}{2 \epsilon_0}{\bf n}$, where ${\bf n}$ is a unit vector pointing from the bottom plate to the top plate. This gives the total ...


0

Charges flow until the potential difference across each capacitor is the same. In each capacitor the product $Ed$ will be the same. The electric field in each capacitor will change because the charge density on the capacitor plates changes.


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Start off by considering what the charge distribution would be like without the battery being connected. The charge distribution must be such that there is no electric field inside either of the plates. The consider which of the charges will change when the battery is connected.


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Your diagram shows that you have almost solved the problem. You have obviously correctly decided on how the charges ($4Q$ and $-2Q$) are distributed on the original capacitor. If you double the area of the plates for the first capacitor what happens to the charge distribution on the plates and what happens to the capacitance of the larger area arrangement ...


0

The trick to this question is that neither situation can exist in reality. There are no perfect insulators (required to isolate the plates in part 1), and no perfect conductors (required for the wires). Both of these situations are ideal cases that exist at the end of an extrapolation of what can happen in reality. To bring this situation back to reality, ...


2

You can't have strong enough electric fields to tear the proton away from the nucleus but it is really a very subtle thing and the inability is just "by a little bit". The strongest electric field that may exist is given by the Schwinger limit. In $\hbar=c=1$ units, the field is $m_e^2 /q_e$. Once you reach this value, electron-positron pairs start to be ...


0

The description of the system is not very clear. I'm assuming that the x-axis is the symmetry axis of the cylinder and that the cylinder extends from x=-L/2 to x=L/2. I'm also assuming that there is a uniform surface charge density on the shell, not a uniform volume charge density over the entire cylinder. Q1 hint: use symmetry arguments for the direction ...


2

A dipole has two parameters magnitude and orientation. If you change the orientation the nature of the force also changes (i.e from attractive to repulsive), so what you are saying is if you drop a body on suppose say North Pole and it is free falling towards earth then on the South Pole it should free fall away from earth which is not the case. ...


0

The orbit will change. To be concrete, lets say that the satellite and planet are illuminated by ultraviolet light for a short period of time, so that some electrons are photoemitted from their surfaces. Make the bodies conducting so that the surface charge spreads out evenly. The repulsion between the charged bodies acts in the opposite direction of the ...


1

The answer to this question is a lot like the answer to Why does the comb attract the pieces of papers if they're neutral? I'm guessing that your comb, which will be negatively charged, is not evenly charged across its width. So even though you are holding the center of the comb above the vane, the electric field between the comb and the ground (plane) ...


0

The four (metal) vanes of a Crooke's are attached to an axle. There is very little friction between the axle and its supports so a very small torque applied to the vanes would produce a noticeable change in the rotation of the vanes. My suggestion is that charges are induced on the vanes by the charged comb. Thus there is a net force of attraction between ...


0

When you choose an arbitrary gaussian the flux is still given by the charge inside the surface. However it would be useless in calculating the electric field unless you are able to write $$\int\vec E\cdot d\vec A=\int EdA=E\int dA.$$ That is, Gauss law is useful when the surfaces elements are always parallel, antiparallel or perpendicular to the electric ...


0

You would calculate the flux $\Phi$ as \begin{equation} \Phi=\int \vec{E}\cdot d\vec{A}, \end{equation} just as you would for a sphere or a cylinder. Gauss's law is true for any closed surface. Just take your electric field and take the dot product with the differential area $d\vec{A}$ of the surface and integrate this over the entire surface. The tricky ...


0

If you argue that capacitors in parallel have the same p.d. across them and in series otherwise, then when the switch is thrown the capacitors are in series (the capacitors have different pd's) and the charged capacitor acts like a source of emf. After some finite time (assuming some finite residual resistance) afterwards the capacitors have the same pd so ...


1

The mistake you made was essentially transforming the force twice. In the frame in which the matter is stationary (I'll call this the primed frame), you correctly found: $F'=q\sigma'/2\epsilon_0$ and in the frame in which the matter is moving (unprimed) you correctly found: $F=q\sigma/2\epsilon_0 \gamma^2$ Since $\sigma'=\sigma/\gamma$ this is: ...


0

simple answer: They are in parallel when both blades of the knife switch are closed. If only one blade was closed, the 2 capacitors would be in series between the 2 open points.


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Similar problem is two charged particles $q_1$ and $q_2$ separated by the distance d. The TOTAL energy is $E=q_1 \ f_2 = \frac{q_1 \ q_2}{d}$, where $f_2$ is potential of the second particle. Thus, you see that you take the charge of the first particle and multiply by the potential of the second one. It gives you the total energy of system.


2

The RHS side of Gauss's Law, that is the charge enclosed should remain the same is indeed true. The apparent confusion if any, should be in the LHS of the equation, the integral of the 'dot product' of field and area vectors. Consider the diagram, Now, when we take the dot product of the field vector with the area vector in the initial case, the field and ...


0

You first need to know that how the current flows through a capacitor. There are two parallel plates with a dielectric between them such as air etc. One plate is connected to the positive terminal of the battery while other is connected to the negative terminal of the battery. Charges from the negative terminal will accumulate on the plate connected to it. ...


0

Electrostatics is the physics of current free charge distributions. Magnetostatics is the physics of stationary (time independent) current distributions. Usually magnetostatics is defined as the physics of stationary and "divergence free" current distributions, however, a zero divergence is a superfluous condition that is not satisfied in case of many ...


0

Because potential exactly means that you have its potential at every point and to get from start to finish you need to make a number of steps. That is one leap can be broken into a series of steps, every step is closer to the target. As you move closer to the target, your potential becomes closer to the target potential. That is, you have some potential ...


0

Q1. But how this can be explained "theoretically"? I assume your question is about the concept of 'electric potential ' due to a distribution of charges and in the present case 'a dipole'. The best way is to imagine an unit positive charge being carried/moved from infinity to a point on the equatorial line of the dipole. Naturally your probe charge ...


0

If you look at the electric filed lines you will note that a positive charge moving in the direction of the blue arrow does have a force on it but that force is always at right angles to its motion. Hence no work is done moving the charge.


0

There is no limit on the 'ability to store charge' involved. What is different, is the proportional relationship between stored energy per unit charge (voltage) and amount of charge (ampere-seconds) stored. A rechargeable battery keeps its charge chemically bonded (and stores a LOT of charge for a given voltage rise), while a close-spaced-plate capacitor ...


1

The diagram above has a very important feature. It is the connection between the Earth and the outer conducting shell. Assume that the Earth is a conducting sphere and has some net positive charge on it. This will mean that the outer shell connected to it will also have some positive charge on it but the wire between the outer shell and the Earth means that ...


1

You are confusing chemistry and physics. In chemistry you may have learnt that when metals react, they give up electrons (the are oxidized). That is because there are usually a small number of electrons in the outermost orbit of the atom, and when these are released the atom is left with a very stable electron configuration. But when you are looking at a ...


0

So, the potential is not necessarily zero both at the ground and at infinity. First off, there's nothing in physics which forces the potential to be zero at infinity; it just happens to be a nice way to think of the Coulomb potential. Potential energies are not absolute numbers; there is a constant of integration that enters into them! Second, there is ...


2

$\nabla \cdot \vec E(r) = \dfrac {1}{r^2} \dfrac {d(r^2 E)}{dr} \ne \dfrac{dE}{dr}$ in spherical coordinates.


1

The differential and integral forms should in principle always lead to the same result, since they are related to each other via Gauss's theorem. (e.g. see What are the differences between the differential and integral forms of (e.g. Maxwell's) equations? ). In this case you have not applied the differential form correctly, because you have used an ...


0

In the situation you depicted, the electric field is different from $0$ in the hollow region enclosed by the conductor and equal to $0$ inside the volume of the conductor (at equilibrium). Why is $\vec E = 0$ inside the volume of the conductor at equilibrium? We will proceed by reductio ad absurdum. Since we are at equilibrium by hypothesis, there can ...


1

I have found your question and the diagram a little difficult to interpret. I have redrawn you diagram to show a charge of $+Q$ on the outer shell and a charge of $-Q$ at the centre together with two conducting shells shaded grey. What else the electric field inside the conductors is zero. If there was an electric field then the mobile charge carrier ...


0

Ok, so it seems my reasoning above was correct as my teacher passed my lab report. If the conductor is divided into discrete regions of different (uniform) conductivity each region will have zero charge density, however at the continuity breaks in conductivity there will be a surface charge and thus the conductor will in fact have some charge density. I ...


0

Car tyres, which these days are almost always radials, are sometimes reinforced with steel. It may be possible that these have become magnetized due to impact in the Earth's field, although I find that unlikely unless you constantly drive in one direction. Otherwise, maybe someone has run a magnet over your tyres? Adding quotes from another paper on the ...


2

If you ignore the coefficients, your function will be: f(x) = x / (x2 + r2)3/2 differentiating w.r.t x will give you: simple product rule f(x) = u(x)*v(x); u(x) = x; v(x) = 1 / (x2 + r2)3/2 f'(x) = u'(x)*v(x) + u(x)*v'(x) f'(x) = [1 / (x2 + r2)3/2] + [(-3/2)2x2 / (x2 + r2)5/2] to maximize/minimize you substitute f(x) = 0: 1 / (x2 + r2)3/2 = 3x2 / (x2 ...


1

What you have forgotten is that $u$ is not a potential rather it is a number with no units. I do not know how you derived your expression for $u$ but here is a way without a lot of the intermediate steps. The arrangement you have described is similar to that of two parallel line charges with separation $2a$ and charge per unit length $\pm \lambda$ as shown ...


1

Using Gauss' law and the condition that there can be no electric field inside a conductor the initial charge distribution is as follows: $+2Q$ charge on the outside of the inner sphere. $-2Q$ charge on the inside of the outer sphere ($-Q$ original charge and $-Q$ induced charge) $+Q$ induced charge on the outside of the outer sphere There is an electric ...


-1

The more simple answer, I suspect is: you're looking at the electric field from a point charge ($\sim r^{-2}$) and not the potential, which goes as $r^{-1}$. If you simply write up your formula with potentials instead of electric fields, the answer is immediate.


2

He meant that the "energy" function $$ E_\textrm{art}(C) = - \frac{1}{2}CV^2 $$ is introduced solely for the purpose of getting the right result with the "principle of virtual work". It is not really EM energy in the usual sense as energy stored in the capacitor, available for use. If there is potential difference $V$, the latter energy $E$ is actually ...


0

It depends on how big your plates are, in the case of two infinite plates the charge wouldn't move. If the plates are finite, the electric field would 'leak' away and the charge would move to the middle and away. In that case there would be a voltage difference, because the voltage near the plates would be positive and the voltage some distance away would be ...


1

This is a homework-like question, so I will not provide a full answer. Here are a couple of good things to think about on your way to the answer: What is special about the velocities (or momenta or kinetic energies) of the particles at the instant of minimum separation? What quantities are conserved throughout the interaction?


1

You haven't seen these effects because your eyes are rather insensitive to wavelength changes. A moderately resolving spectrometer can detect these changes quite easily. It's called "Stark effect" and it can be observed in atomic spectroscopy: https://en.wikipedia.org/wiki/Stark_effect. For magnetic fields the analog is called "Zeeman effect": ...


0

The electric field inside a conductor must be zero so the net charge within the Gaussian surface must be zero. Thus the charge $+Q$ must induce a charge of $-Q$ on the inside of the conductor. If the conductor was originally uncharged then this in turn must lead to the conclusion that the charge induced on the outside of the conductor must be $+Q$. This ...


1

With the capacitor remaining connected to the battery, then the potential difference between the plates is unchanged. If the potential difference stays the same and the gap between the plates stays the same, then the electric field stays the same. Your initial assumption that the field is reduced by a factor $K$ is only true if the capacitor is disconnected ...


3

The flux lines around a point charge are spherically symmetric, so the total flux through a surface is proportional to the solid angle subtended by the surface. When you see questions like the one you cite they usually have some trick using symmetry. For example for the charge in the centre of the face the solid angle subtended by the whole cube is $2\pi$ ...


-1

If apply in moving charge then no. Of forces act on them so we can't calculate position of charges but in case of static it is possible so it applicable only for point charge or static


2

Imagine that the electric field in the region of the neutral (null) point looks something like this with a Gaussian surface which is a cylinder shown in red. The exact shape does not matter. As drawn all looks well in that grey electric field lines enter at faces $A$ and $B$ and leave out of curved area $C$. One can imagine that all the sums work out as ...


3

You forgot to draw two exitting field lines. It's a hyperbolic singular point. It looks like this: Two in along the horizontal line, two out along the vertical line. You seem to have a mathematical problem with singular points. Sure... if you follow along the EXACT horizontal line, sure... you hit the point and don't know how to continue. But that's true ...



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