New answers tagged

0

There are many many misconceptions tied into knots in your question. Firstly, the electric force between two charges doesn't depend on just the distance between them and their charges, it also depends on their velocity. When charge A moves then its electric field is different, so the electric force charge B feels is different. This means you can't reason ...


0

The quarks have fractional charge not only because of fitting proton and neutron constituents. There is a huge data base of hadronic resonances that gave rise to to the quark model , the standard SU(3)xSU(2)x(U(1) of particle physics. The quantum numbers assigned to the quarks are important, among them the 1/3 and 2/3 charge and the color assignements. The ...


0

http://physics.stackexchange.com/a/65392/101895 Brilliantly explained there, relate the contracting effect with classical sound Doppler effect if you could. Again, only force acting on a particle due to another particle is solely coulomb's force.The most important thing is that concept of exchange of particles is supposed to create the force {basically, a ...


0

Yes. If you have two positively charged plates of different charge magnitudes and fire a stream of electrons between them, the electrons will deflect toward the "more positive" plate. Though really the electron stream isn't exactly necessary, since the two plates will repel each other indicating a potential difference right off the bat.


0

If by magnitude difference you mean Plate A has a charge of +1 and plate B has a charge of +9, then there will be a potential difference. If by magnitude difference you mean Plate A has a charge of +1 and plate B has a charge of +1, then there won't be a potential difference.


1

It all depends by what you mean by the word "flow". Let the charged body which is assumed to be a conductor produce an E-field. In a conductor which has mobile charge carriers then the charges can be made to flow within a conducting body which has no net charge. If you subject an uncharged conducting body to an external E-field then the mobile charge ...


3

It sounds like you're asking "If two conductive materials are brought in contact, and one of them is electrically neutral, and one is positively charged, which direction will charge flow?" If that is indeed your question, then the answer is that negative charges (electrons) will flow from the neutral object to the positive object until they are at the same ...


2

The net effect of the charging process is the movement of electron from one plate which then has a net positive charge to the other plate which then has a net negative charge. The battery facilitates this by creating an electric field in the wires and it is this electric field which applies forces on the electrons which makes them move. The movement of ...


2

You are correct, electric current consists of electrons travelling from one place to another. Some materials conduct electricity better than others. Copper is one of the best and that's why our conductors are usually made of copper. Aluminium is also very good (so is silver) and high-voltage cables are usually made of aluminium. However, everything conducts ...


0

Ground is at 0 potential,so,it accepts electron from negative terminal. And at very far place any positively charged electrode accepts electron from ground and current flow.


0

The last part is possibly the most interesting in that you have think about what happens to a dielectric when it is placed in an external electric field, or put another way; how does the movement of charges within the dielectric change the net electric field between the plates?


0

You have the right equation. Since you are not given absolute values, you should just reference your answers to the original $Q$. For example, you can say for the first part "The charge will be 0 when the voltage difference is 0. As the voltage difference increases to $\Delta V$, the charge will increase (proportionally) to $Q$". No numbers were needed... ...


2

The naive reasoning which leads to the conclusion that charges $Q_1$ and $Q_2$ of two touching conducting spheres with radii $R_1$ and $R_2$ are related by the relation $Q_1 = Q_2\frac{R_1}{R_2}$ is wrong. This formula holds only when the distance between the spheres $L$ is large compared to $R1$ and $R_2$, $L\gg R_1,R_2$, and the spheres are connected by a ...


-2

Its simple, electrons are on the outside of the nucleus of the atom. While protons and electrons can be transferred, they are on the inside of the nucleus and if they were transferred it would be considered a nuclear reaction.


4

In classical electrodynamics, assuming a point charge to be having a finite charge, the net electrostatic self energy carried by it is given by $$ Self Energy = 1/2 \int E^2 dV$$ Upon performing the intergral in three dimensions, since the electric field of a point charge diverges at the origin, therefore the rest mass by the virtue of the electrostatic ...


1

Look if you substitute the boundary conditions in the given solution it will be identically satisfied, so you don't need them since those were used in the derivation of the given solution. So you have found the charge density and this should be satisfactory, though the final result should be: $$ \rho=-\frac{\rho_0}{\epsilon_0}\cos\beta x , $$ as pointed out ...


1

The direction is correct, just note that you can write $$ \frac{\partial^2\phi}{\partial x^2} = - \beta^2\phi $$ and $$ \frac{\partial^2\phi}{\partial y^2} = \beta^2\cdot\left( \phi -\frac{\rho_0}{\epsilon_0\beta^2}\cos{\beta x}\right) $$ This can ease the calculations. The result you obtain is an oscillating potential in $x$, not depending on $y$. The ...


0

You're correct. And you're not. Depending on how you read the question. There's more than one way to interpret the phrase "the force exerted by q3 on q1." You've interpreted it as just one in a bunch of terms in a net force equation, such that the net force on q1 is q1 times the sum of all the electric fields produced by all the charges in the problem. One ...


0

Assuming the cylinders are conducting. The method of images is derived by showing that the equipotentials of two lines of opposite charge are cylinders or a plane. You then have to do a bit of algebra to figure out where to put the charges so the equipotentials coincide with the given cylinders. (The charge will not be at the centre of the cylinder and the ...


2

q1 and q2 induce charges of the same magnitude and of the opposite sign on the surface of the spherical cavities. This in turn means that on the outside surface of the sphere there are charges induced of the same magnitude and the same sign as charges q1 and q2. These charges are distributed on the outside surface of the spherical conductor. That is there ...


2

In electrostatics, we generally assume our conductors to be ideal. This indirectly assumes that charges have free mobility inside the conductor. You must remember that a system is more stable the lower its energy is. A system of free charges always tries to assume a configuration in which its potential is the lowest. (This configuration is achieved because ...


0

Sign conventions aren't "wrong", but they can be misleading. For example, we could re-define work done in a gravitational field so that escaping earth's gravity well would require negative work. That's an equally valid convention, but we associate positive work with effort, so reversing the convention would hinder our physical intuition for no discernible ...


1

In quantum field theory, particles are simply excitations of fields. And interactions are determined by symmetry in an extremely elegant way, see gauge principle. Symmetry is the central concept in fundamental physics. Except that it determines the interaction, it can be also used to classify particles. For instance the spin of particles is characterized by ...


0

The weapon you are suggesting is similar to one where a laser pulse ionizes the air to the target and channels a high voltage high current charge to the target. It requires extremely high power laser pulses in the femtosecond region. The name of the tech is Electrolaser


-1

This weapon exists and is commercially available. The amount of heat discharged is minimal, but it's excellent at disrupting electrical oscillators like hearts and low-current electrical pathways like nerves.


0

Not at all. The split of the 6 component electromagnetic field into a 3 component electric field and a three component magnetic field is an artifact of choosing an inertial frame. A different frame will break the same electromagnetic field into different parts. Just like if you have combinations of jumping to the future and teleporting in space then when ...


1

Although the eV is defined with respect to a particle of unit charge (an electron) in an electric field, it is simply a unit of energy. There are simple relations between thr eV and everyday units of energy, such as the Joule or calorie. Thus energies of all sorts of things, in fact any energy, can be expressed in eV, even if it has nothing whatsoever to ...


1

One electron-volt $=1.6 \times 10^{-19} joules$ and is a unit of energy that is equal to the energy acquired by an electron falling across a 1 volt potential difference. The particle (neutrino) doesn't need a charge to have some energy. Instead of expressing the mass of a particle in kg, we can express it as $mc^2$ which is an energy (joules or eV... your ...


0

No, this is not true all the times for every region in space. Take this example: If you have two charges with the negative charge. At the middle of the straight line that connects them, the electric field(the force) will be equal to zero because the force from the electric field of each charge will be equal in magnitude but they will have opposite ...


1

The effects of gravity are really only observable to us on a macroscopic (large) scale. When a large enough number of (perfectly neutral) Hydrogen atoms come together they will gravitate towards each other. That sets things in motion for the Hydrogen to heat up. Once they reach a high enough temperature and density, they will ionize and the protons can ...


-1

You can read lecture notes by Malcolm Perry on the Applications of Differential Geometry to Physics. A detailed answer to your question is given in pages 37 & 38. I am providing the link for the (unofficial) lecture notes http://www.aei.mpg.de/~gielen/diffgeo.pdf


3

The Kaluza-Klein equations of motion (the geodesic equations) for a particle moving in the 5D spacetime contain the equations of motion of a particle in 4D spacetime under influence of electromagnetism if and only if one identifies $p^5 = mU^5 = \frac{1}{\sqrt{G}}cq$, i.e. relates the momentum in the fifth dimension $p^5$ to electric charge $q$. (And yes, ...


1

I think you use "relative vs absolute" to mean "distinguishable vs undetermined". If this is the case, we could say it is a possibility that yes, the charge of neutrino is undetermined (relative). This is because having a charge, is physics jargon for "susceptible of certain kind of interaction". Thus neutrinos have no electric charge (do not "feel" ...


0

It really depends on what you mean exactly by "charge." A neutrino has no electric charge, so we'll never find an electrically charged particle that interacts electrically with it. On the other hand, neutrinos (and other particles) have other intrinsic properties that really only differ from charge in their function. For example, neutrinos have mass, so ...


2

This is too long for a comment. From Wikipedia In the most recent CODATA adjustments, the elementary charge is not an independently defined quantity. Instead, a value is derived from the relation $$e^2 = \frac{2h \alpha}{\mu_0 c} = 2h \alpha \epsilon_0 c$$ where $h$ is the Planck constant, $α$ is the fine structure constant, $μ_0$ is the ...


1

On the face of it, the answer is "nothing will happen". However, if you bring the surfaces close enough together, you may find that electron affinity between the two is different, in which case electrons may move by a very small amount - in the same way that when atoms react, the resulting molecule may have a dipole moment. The effect would be restricted to ...


0

The potential difference across them wouldn't be equal because their lengths are equal. It would only be equal if their resistances were equal, which as you point out they are not.


0

If they are in series, then yes the current will be the same flowing through each, it has no other path to take.



Top 50 recent answers are included