# Tag Info

1

This sort of calculation, especially when the speed of the electrons from an observer's point of view is close to $c$, has to be done using special relativity, in that sense that the transformation between reference frames is determined by Lorentz rather than Galileo transformations. As you mentioned, you can put your reference frame origin at one of the ...

1

Boundary conditions of electromagnetic waves are well looked into because of the nature between polarization and reflection. However in this particular case it is rather simple, it is the evaluation of a contour integral over the boundary. This is done by summing the length elements of the pill-box shape that is typically used in this analysis in the correct ...

1

You just use vector addition and Newton's Second Law. For example, if you have $$\overrightarrow{F}_E=qE\hat{x},\space\space\space\overrightarrow{F}_B=qvB\hat{y}$$ then your total force $F_{tot}=F_E+F_B$ is just $$\overrightarrow{F}_{tot}=qE\hat{x}+qvB\hat{y}$$ Since $\hat{x}$ and $\hat{y}$ are totally linearly independent, these terms cannot be combined. ...

0

Consider a metal sphere of unit positive charge. Now , place a point charge at centre of sphere. What do you observe? You find that electric field inside sphere is not zero. Why? Because , electric field due to charge on metal is cancelled and not by charge inside metal sphere. Returning to your question Similarly, Electric field due to ...

0

I assume you are talking about linear materials with dielectric constant greater than $1$. Say you have a free charge distribution $\rho$ in vacuum, which produces an E field. Now you introduce linear material. Then the free charges will be "weakened" because they will by partially screened by charges of the dipoles sticking on them. However, the opposite ...

0

(For completely filled capacitors) Q = CV So, C = Q/V So, C is charge stored per unit Potential Difference applied. Now, V = Ed ,where d is distance between plates. $E = \dfrac{V}{d}$ Case 1) When you apply a constant V of 1V to capacitor E across capacitor is $\dfrac{1V}{d}$ which is constant independent of capacitance of capacitor or dielectric b/w ...

0

For a discrete system of $N$ charges, the potential energy associated with their configuration is given by \begin{align}U&= \frac12\,\sum_{j=1}^N \, q_j\sum_{k\ne j}\,\frac1{4\pi\epsilon_0}\,\cdot \frac{q_k}{r_{jk}}\\ &= \frac12\,\sum_{j=1}^N \, q_j\,\varphi(\mathbf r_j)\tag 1 \end{align} where $\varphi$ is the scalar potential due to all charges ...

0

You have $${U = \frac{1}{2}\frac{Q^2}{\frac{\epsilon_{0}A}{d}}}$$ $${\frac{\epsilon_{0}A}{d}}=C\equiv\textrm{ the capacitance of the capacitor.}$$ Hence $${U=\frac{1}{2}\frac{Q^2}{C}}$$ Now $\displaystyle{Q=CV}$, where V is the p.d. between the capacitor plates. So, $${U=\frac{1}{2}CV^2}$$ or $${U=\frac{1}{2}\frac{\epsilon_{0}A}{d}V^2}$$ ...

0

If it wouldn't be its tangential component would exert a force on the charges and they would move. The condition would then not be static. After some movement and redistribution of charges (when there would be no force on charges) the condition will again become static thus making the field only normal to the surface

3

As stated by Lemon, electric flux through a volume enclosed by a closed surface is zero when the volume contains no net charge. Electric flux through a closed surface $\rm S$ is $$\Phi= \int_{\mathrm S} \,\mathbf E\cdot \mathbf n\,\mathrm d^2 \mathbf r\;.$$ Now, according to Divergence Theorem, \begin{align}\int_{\mathrm S} \,\mathbf E\cdot \mathbf ...

1

For a charge distribution $\rho({\bf r'})$, the electric field at ${\bf r}$ is $${\bf E}({\bf r})=k\iiint\frac{\rho({\bf r'})}{|{\bf r}-{\bf r'}|^2}\hat{\bf R}d^3{\bf r'}$$ where ${\bf R}={\bf r}-{\bf r'}$. I think the OP's claim is at positions where $\rho \ne 0$, the above integral is infinite because the integrand blows up at ${\bf R}={\bf 0}$. I think ...

2

The equation you provided is actually given by $$W=\frac{\epsilon_0}{2}\int E^2\,\mathrm d\tau$$ which is the energy stored in an electric field. This energy is utilized by the charge to generate it's field of influence or it's electric field. It's dependent on the magnitude of charge and the distance of separation between the charge and the point of ...

3

A field is a field: not a force. A force describes the effect of the field on a specific "object" (see below) coupling to the field. In the current state of research, quantum fields (such as the photon field for electromagnetism, see Virtual photon description of B and E fields) are fundamental structures that cannot be decomposed or explained in simpler ...

1

Your thinking is correct. At least, if the sphere is made up of small point charges, then the field will be infinite as you approach them. There is a point you are missing when you say that this contradicts Gauss' law: Gauss' law only gives you the flux of the field. To get the field of the sphere from it in the textbook-way, you have to use symmetry. You ...

0

I) The Dirac delta distribution (and derivative thereof) in the dipole field $$\Phi ~=~\frac{1}{4\pi\varepsilon}\frac{\vec{p}\cdot \vec{r}}{r^3} \tag{1}$$ $$\Downarrow$$ $$\vec{E}~=~-\vec{\nabla}\Phi ~=~ \frac{1}{4\pi\varepsilon}\frac{3(\vec{p}\cdot \vec{r})\vec{r}-r^2\vec{p} }{r^5} -\frac{\vec{p}}{3\varepsilon}\delta^3(\vec{r}) \tag{2}$$ $$\Downarrow$$ ...

0

There are two equivalent definitions for the potential difference between two points. The potential difference between two points is the work done by an external force in moving unit positive charge from one point to the other. The potential difference between two points is the minus work done by the electric field in moving unit positive charge from one ...

0

"The divergence of any electric field created by any surface charge distribution would be zero." No it's not. Consider a charged conducting sphere with uniform surface charge density and a Gaussian sphere of radius greater than the original one. The electric field is diverging through the surface of the Gaussian sphere. So the divergence cannot be zero. ...

15

The fact is that, in the general case $$\vec{E} = -\vec{\nabla}V - \frac{\partial\vec{A}}{\partial t};$$ (signs depend on conventions used) where $\vec{A}$ is called vector potential. You can consult for example Wikipedia. Let us consider homogeneous Maxwell equations: $$\begin{cases} \vec{\nabla}\cdot\vec{B} = 0,\\ \vec{\nabla}\times\vec{E} + ... 2 For dynamic electric and magnetic fields, there is a piece of the electric field that depends on the vector potential:$$ \vec{E} = - \vec{\nabla} V - \frac{\partial \vec{A}}{\partial t}, \qquad \vec{B} = \vec{\nabla} \times \vec{A}. $$Taking the curl of the first equation yields Faraday's Law (with the V-dependent term dropping out as you note); taking ... 6 When there is a time-varying magnetic field, the electric field is non-conservative and therefore cannot be written in the form \mathbf{E}=-\nabla V. 0 Well, your statement is incomplete. Electrostatic field is zero inside the MATERIAL of the conductor. It may have any other value inside( inside does not include the material of the conductor) or outside the conductor. It is zero inside the MATERIAL of the conductor, because since charges are free to move, they generate a electric field due to polarised ... 0 All the points on the inner surface of the hollow conductor have the same potential. Therefore, if the charge density \rho inside is zero everywhere, the electrostatic field will be zero. In your case the filament carries a charge Q, so the charge density is not zero everywhere. 0 Dielectric constant K is actually the same thing as relative permittivity, and it increases the overall permittivity \epsilon. So in general, whenever you see the permittivity of free space \epsilon_0 in an equation, if you're dealing with a dielectric, you can multiply it by the dielectric constant and see how the equation changes. For example, since ... 1 You will get infinity because in addition to kQ_1Q_2/d, it also includes the self-energy of the two point charges, which is infinity. 0 Yes, because electric field lines travel at the speed of light in vacuum.However its effect at infinite distance would be negligible, as if field did not reach there at all. 1 The range of the Coulomb force is infinite (the force between two charges Q_1, Q_2 separated by a distance r is given as F = \frac{Q_1 \, Q_2}{4 \pi \epsilon_0 r^2}), with the implication that the photon has zero (rest) mass. However if you were to suddenly create (say) a positive then the "news" about this would travel at the speed of light, so that ... 1 When doing electrodynamics, you don't really consider a finite distance at which field lines extend - you consider the behavior of the field as it extends to infinity. In other words, when you have a potential that's dependent upon a distance r from the origin, you take the limit as$$ \lim \phi (\textbf{r})\rightarrow\infty $$and see what happens (e.g. it ... 0 To place a charge in the vicinity of an electric field, you should do work against the electrostatic force on the charge. This work done to bring a charge q to an electric field of some other charge configuration from infinity to a distance r, in the field is what we call the potential at the point r. To do a work to move a charge q from a potential V to a ... 1 Relation between Electric field and potential The relationship between electric field \bf E and scalar potential \varphi is given as$$\mathbf E= -\mathbf \nabla\,\varphi$$where \mathbf \nabla \equiv \textrm{gradient operator}\;. I am unable to understand from this - sign comes. It is worthy to quote from Purcell: The minus sign came in ... 1 I guess you mean$$V=\frac{1}{r}$$instead of V=1/x (if \partial_y V=\partial_z V=0 all your integrals diverge). If this is right, then$$V=\frac{1}{r}\Longleftrightarrow \rho\propto \delta(\boldsymbol r)$$and not \rho\propto \frac{1}{r^3}. Anyway, your example V=1/r is highly pathological: it diverges too fast at the origin (the integrals blow up ... -1 An electric field is the force that fills the space around every electric charge or group of charges. Electric fields are caused by electrical forces. Electrical forces are similar to gravitational forces in that they act between things that are not in contact with each other. Electric fields are also analogous to magnetic fields resulting from forces acting ... 0 The charges could equally well be the other way round and then the direction of the electric field would be reversed. Perhaps the negative charges where placed on the inner conductor and an equal number of positive charges were induced on the inside of the outer conductor? There will be a potential difference across the conductors due to an excess of ... -1 At r=R, the field strength will be maximum and the field decreases as you go away from it. But that's not the case when you go inside. Any point inside the sphere contains no electric field. The field suffers a discontinuity at the surface. But however the potential is a constant through out inside. It makes sense. The potential is continuous. The potential ... 0 Those formulas, and those graphs are idealizations. In reality, there are no discontinuous fields, just as there are no zero-thickness shells. If you start with an impossible situation, you will calculate impossible and meaningless results. For real situations, situations for which the theory is valid, the field may change quickly, but it does so ... 0 Electric field lines are always at right angles to equipotential lines or surfaces. The electric field is minus the potential gradient. So in the diagram showing a uniform electric field a positive charge would experience a downward force in the direction of decreasing electric potential. In this case the magnitude of the electric field is \frac ... 0 The electrical field is related to a force concept: it describes the force per unit charge. The potential is related to a potential energy concept: it's the added electrical potential energy per unit charge. So, just as the force is the negative gradient (or in 1-dimension, the negative slope) of the potential energy function, the electric field is the ... 1 how the direction of the force on slab in both situation differs? Recall that the energy stored in a capacitor is given by$$W = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C}$$where V is the voltage across the capacitor and Q is the magnitude of the electric charge on either plate. For the case that a constant voltage source (e.g., battery) is ... 0 We defined positive and negative charges because it allows a very convenient and accurate physical representation of charges, with relative numbers. The choice of ‘who is called who’ is totally arbitrary to begin with. We could have called positive charges ‘negative’ and vice-versa. We also could have called them « blue » and « red » (actually, I think ... 0 If the grey and blue parts of your diagram are the conductors and the magnitudes of the two sets of charges are the same then you have drawn a correct diagram with an electric field present only in the region between the two sets of charges. The field in that region being the same as if there was a -Q charge at the centre of the arrangement. In all other ... 0 If you are talking of a conducting sphere, this situation cannot happen physically. Charge is supposed to stay on the outer surface of the conducting shell. However if you make the shell infinitely thin (which is again not possible physically,) then the potential inside (for radii less than the radius of the shell) would be still a constant. 0 You could just calculate the vector components. These hand rules are used whenever vector quantities are related by a cross product. Let's take as an example the force on a moving charge due to a magnetic field:$$\vec{F}=q \vec{v} \times \vec{B}$$Let's write this out in terms of its components (assuming we're working in 3D):$$\left( \begin{array}{} F_x ...

2

A closed surface like a sphere encloses some volume. Anything coming out through the surface (the net outward flow which we call the flux) will be in the expense of what remains inside. If the sphere encloses some charge, then electric field diverging out from the volume containing the charge will be equal to the normal component of the electric field lines ...

0

Not all vector fields have zero flux over any closed surface. Those having zero flux everywhere are a special type called the 'solenoidal'. The electrostatic field doesn't belong to this group. I think you need to study some vector analysis to capture the concepts correctly.

0

Grounding doesn't set the charge to zero - it sets the potential to zero. That means the residual charge will be whatever it needs to be to achieve that.

0

The potential energy is related to a system not to an individual charge. So when you bring one charge towards another charge the system of two charges gains potential energy. What this means is that any work which is done by either or both of the charges will result in the loss of potential energy of the system (both charges).

0

In your first case you said you were bringing a charge B to the first charge lets say A. Here your reference is A thats why u say you are bringing it to A. Think of it the other way, put you reference on B then your perception would be you are bringing A to B and now A will have a Potential energy. Thats how it works both ways. And Finally i would like to ...

1

The potential energy possessed by the system of two charges is $$U=\dfrac{Q_1Q_2}{4\pi\varepsilon_0r}$$ You can rearrange the equation into$$U=\frac{1}{2}\dfrac{Q_2}{4\pi\varepsilon_0r}Q_1+\frac{1}{2}\dfrac{Q_1}{4\pi\varepsilon_0r}Q_2$$ which then becomes$$U=\frac{1}{2}Q_1V_1+\frac{1}{2}Q_2V_2$$ where $V_1$ and $V_2$ are respectively the potentials created ...

1

The electrons follow the field as far as they may and as far as there is any field. Be aware, that there is no such thing as "positive field" as you wrote. The field is a vector and has a direction. Consider first an easier setup: if the external charge would be inside the conductor, the electrons would come and surround it and cancel it completely, there ...

2

Because the potential is a word that describes how much charges want to move somewhere else. They will always want to move to the lowest possible potential. If you have high potential in one conductor, then it means that there is a lot of equal excess charge in it. These charges repell each other, so of they were allowed they would move away. If you have ...

2

Since, as long as there is a potential difference there is an electric field in the direction of higher to lower potential. $$\textbf E = -\nabla V \qquad \qquad \textbf F = q \textbf E$$ The charges keep flowing from higher potential region to a lower potential region due to the force of this electric field. The accumulation of charge in the lower ...

Top 50 recent answers are included