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HINT- Split the dipole into into components, such that one of the components of the field at the equipotential surface is axial and one is equatorial. Draw these along with the applied field at that point. Since the surface is equipotential, the net tangential field must be zero. So, write $\vec E_{tangential}$=0. You should get something like $(\frac{p}{4\... 1 This looks like a homework problem, so I'll just give some ideas on how to start this problem. The electric potential of a point dipole$\vec{p} = p \hat{z}$can be written as $$V_\text{dip} = \frac{1}{4 \pi \epsilon_0} \frac{\vec{p} \cdot \hat{r}}{r^2} = \frac{p}{4 \pi \epsilon_0} \frac{\hat{z} \cdot \vec{r}}{r^3}.$$ The electric potential of a pair of ... 2 I sort of doubt the blue jet explanation by Xeren is probable (although it is a possibility). It's just too rare and too faint. Much more likely, it's lightning far away near the horizon. Just because there are no clouds doesn't mean the light wont scatter. Sunlight makes the sky very bright blue during the day, and in the brief moment when lightning ... 2 Brief Summary Numerically, the mean value property of harmonic functions allows you to get an approximate solution to boundary value problems relatively quickly. Often you can improve convergence to a solution with a good initial guess, however, so analytical approaches can still be useful. Consider the limit of an infinitely long cylinder. There is a ... 1 It's just that the sum of kinetic and potential energies are constant through all times. You can say that potential energy is zero when they are far away, so when they are approaching, the potential becomes more negative, as it is converted to kinetic. Or you can say that the potential energy is largely positive when they are far away, then reducing to zero ... 1 Here is the simple high school level answer which knzhou's answer is better than. The electric potential energy of a system of two point charges a distance$r$apart is given by $$E_E=\frac{kQq}{r}$$ If the charges have opposite signs than the potential energy will be a small negative value when they are far apart. As they move closer together potential ... 1 The simple 'first year physics' answer is that the potential energy goes negative. The negative potential energy cancels out the positive kinetic energy, leaving the total energy equal to zero. This might still feel unsatisfying, because it still looks like the kinetic energy is coming 'out of nowhere'. The real resolution is better. In this situation, '... 4 For point sources of a field or energy source, such as a charged particle, a gravitational body (which acts like a point source), or a loudspeaker on top of a tall column, the geometry of the problem controls how energy and fields distribute themselves in space. At all points that are an equal distance from a point source, the energy or field strength is ... 0 I can give you an answer using only symmetry considerations, and leave to you the rigorous calculations, since you are probably familiar with differential calculus. The figure below shows a transverse cut of your system of two shells. For any point-like area of the inner shell (red dot) you can draw a diagram like this, which is symmetrical for rotations ... 1 you are basically trying to undergo a transition from a law which is valid for static charges(or non relativistic speeds) to one which is valid for steady currents. That is why, simple differentiation is erroneous and does not include any magnetic field term in dE/dt. 0 If you are asking for an explanation of the result, it is because the finite disk behaves as a point charge for x>>R. Or are you specifically interested in mathematically proving the result? 1 When you calculated$dE_x$, you multiplied by$\frac{a}{\sqrt{x^2+y^2}}$. Why? Hint: try$\frac{x}{\sqrt{x^2+y^2}}$0 If the dipole is small enough, then the force on dipole would be: $$\vec{F}=\nabla(\vec{p}.\vec{E})$$ and consequently the torque would be: $$\vec{F} \times \vec{r}=\nabla(\vec{p}.\vec{E}) \times \vec{r}$$ where r is the length of the dipole 2 You are not completely mistaken. What you have to do to get the desired formula is a Taylor expansion of the term which contains$R/x$and then consider the limit when$x \to \infty$. Also be careful with notation,$\sigma=Q/ (\pi R^2)$. As Andrea Di Biagio mentions in his comment,$1/(1-u) \approx 1 - u$when$u$is small. In your situation$u=R^2/x^2$. ... 0 The torque$ \tau $on an electric dipole with dipole moment p in a uniform electric field E is given by $$\tau = p \times E$$ where the "X" refers to the vector cross product. Ref: Wikipedia article on electric dipole moment. I will demonstrate that the torque on an ideal (point) dipole on a non-uniform field is given by the same expression. I use ... 0 Gauss law in 2D would have to be: $$\oint \mathbf{E} \cdot \mathbf{\hat{n}} dl = 2 \pi q$$ because you are reducing your surface in 3D to a line in 2D, and keep the idea of measure of the boundary and its orthogonal direction or normal. To get the expression of the field you have to make use of the fact that the electric field is isotropic. In other words,... 0 Deriving the electric field for a 2D world can be done in several ways. It would depend on what behavior of the electrostatic interaction you want to preserve in that world. I you asked Coulomb, when he published his expression for the interaction, he would probably have said that the expression should be the same ($1/r^2$) just that the distance$r$would ... 0 In the expression $$\mathbf{p_{right}} = \sum_i^N q_i\mathbf{d_i}$$ you are summing dipoles and assuming that the net charge is zero so the sum is independent of the choice of reference point. So$q_i$is the charge on the dipole and$\mathbf{d_i} $is the position vector of the positive charge relative to the negative charge. If you have two charges$+q$... 1 You said "Fermi level is constant throughout the junction" - that's correct. But fermi level is "A" (see top right in the table). So A is constant (you can set it to zero if you like). B is not constant. 0 I am still not sure of your problem physically. Mathematically, it seems to be about finding the solution to: $$\nabla^2 \Phi = \frac{J}{\sigma} \delta(\vec{r})$$ with$\Phi(\vec{r}=R\vec{e_r})=0$for$\vec{e_r}$being the boundary condition. And you are right, this can be solved using the method of a Green's function, the one obeying the auxiliary ... 2 The phenomenon you are talking about is called dielectric absorption. The way it works is this: Let's say you've just discharged a capacitor. An ideal capacitor would remain at zero volts after this. However, in real life, the capacitor will develop a small voltage from time-delayed dipole discharging (also known as dielectric relaxation). Dielectric ... 1 An electron gun is used to shoot electrons at the ink which then gives the ink droplets a negative charge, varying based on where the ink needs to go. Then, the charged ink droplet passes between two metal plates, which deflect the ink to its appropriate location on the paper. The ink does not acquire charge from the metal plates, but from the electron gun. ... 0 The plane is an equipotential surface. Since the electric field is the (negative) gradient of the potential, the electric field can have no non-zero component parallel to the surface. 0 Assuming y is perpendicular to the plane, you are correct, by symmetry you cannot have a preferred direction along the plane, so the electric field components must be zero 0 Saare taate fado bhancho The dielectric will develop charges such that the net force in that direction stays the same I.e the along the positive plate the dielectric will develop some negative charges such that the net force remains same 1 An isolated charge (like, an electron) is produced by ionizing an atom, using energy to pull a single electron free of the atom and pulling that charge far from the opposite-charge ion. So, it DOES take energy to isolate the charge. The isolated charge has an E field around it, but the original uncharged atom had none. Similarly, when you apply a ... 0 Even electrostatic fields still contain potential energy in them. The issue here is, an electron (for example) cannot interact with itself. This means that its own field cannot give it a potential energy. So yes, in the case of a lone charge completely isolated, it will create an$\vec E$field which contains potential energy, but it will do no work until ... 0 As per my understanding of dipoles : When we talk about dipole we usually think of a charge pair of +q-- -q separated by a distance of 2d. But we are much far away from the dipole. i.e. 2d << D where D is the distance of observer from dipole. If this condition do not hold the charge dumbbell can not be treated as dipole. You have to treat them as ... 1 Indeed, there must be dipole moments of such magnitude occurring (if you just base it on the definition)! But usually, there is no extensive electric field to support the continuous rotation of the dipoles, because the net electric field in space is zero. Though there will occur some abrupt torque (again, based on definition) of intense magnitude due to the ... 2 The idea that in a dipole, the further you remove something, the stronger the interaction becomes is equally counter-intuitive to me. The dipole moment does not define an 'interaction'. You can just interpret it to be a useful mathematical quantity - nothing more, nothing less. For example, if I removed two charges to infinity from each other, their ... 2 You can always see potential as an 'energy difference' between two points. For example in some physical problem you choose a certain zero point of energy and you take that as a reference point for the other energies in the problem. You can define that the potential energy of a mass is zero when it is at a certain height and then calculate the potential ... 0 The constants in your expressions secretly contain units. You can determine what they are algebraically. For example considering the dimensions (denoted with [square brackets]) of$E_x$: $$\textrm{[units of E_x]} = \textrm{[units of 6]} \textrm{[units of x]} \textrm{[units of y]} ,$$ $$\textrm{[V/m]} = \textrm{[units of 6]} \textrm{[m]} \textrm{[m]}... 3 Imagine you are looking through a telescope that zooms in and only lets you see a very small region of the plane. Let's think about what happens to the electric field from the area you are looking at as you move the plane away from you, while you keep the direction of your telescope fixed. If you double your distance from the plane two things happen, and ... 1 I have found this clear answer to your question: I mean Raziman T.V.'s explanation. Enjoy it! Here's the answer: Feynman explains this well in his lectures (Chapter 13 : Work and Potential energy (A)). The explanation is for the gravitational field of a plane, but the same argument works for the electric field of a charged sheet as well. The ... 3 The detailed answer by NowIGetToLearnWhatAHeadIs (and Andrea) is the correct one and I will leave my answer as it is so it can be viewed by other users who might have thought of this themselves in order to understand the fundamental mistake they make. This answer Implies that this goes for any kind of force field with a source of similar geometry, while the ... 0 The force F between capacitor plates is discussed here and shown to be \frac 12 QE where Q is the charge on the capacitor and E the electric field strength. In your example with a constant voltage this is better written as F=\frac 12 CV \; \frac V d where C is the capacitance =\frac{k\epsilon_o A}{d} with d the separation of the plates and ... 4 The converse is not a "theorem" in that you can't prove that it is true, even assuming Maxwell's equations as axioms. It is simply a "hunch" that |\vec{S}| represents the power intensity and U=\frac{1}{2}\,\epsilon\,|\vec{E}|^2+\frac{1}{2}\,\mu\,|\vec{H}|^2 the energy density. What you can prove (and what you already understand) from Maxwell's equations ... 0 The purpose of a Van de Graaff generator is to produce a source of electric charges which are at a high potential relative to the ground - it is an electrostatic generator. A motor moves electric charges which are on a belt made of an insulator into the inside of the dome which is going to store the charges. Within the dome those charges are taken off the ... 2 This is a more general answer. Atoms and molecules are quantum mechanical entities. This means that the "shape" of atoms depends on the solution of quantum mechanical equations, which give probabilities for locating in space the electrons that are bound to the nucleus of the atom with the electric potential provided by the protons of the nucleus. The ... 5 The water molecule is neutral on overall basis, i.e: the water molecule as a whole has no net charge. The water molecule is not linear rather it has a bent shape with two hydrogens on the same side. This happens because of the lone-pair-bond-pair repulsions. The oxygen has is a more electronegative element than hydrogen, i.e: oxygen has high electron-... -1 As the potential is equal, there will be equal amount of work done by the electrons to move from infinity to any point on the surface. If the surface has different potential, then the electron will be accelerated, then it will automatically become a non equipotential surface. Thats why it should be normal. -1 Note-: My grammar isn't that good. But with the help of diagrams, I hope everyone will understand. Lets consider any closed surface in space with the charge (or origin) anywhere inside it. Lets also consider an infinitesimal solid angle from the origin. Let dS be the infinitesimal area by which an infinitesimal solid angle is subtended.(Since there are ... 1 Often the easiest way to do such problems is using potential rather than the electric field. And always, it is best to use the appropriate coordinates. In this case, cylindrical coordinates with the rod at \rho=0 and extending from z=-L/2 to z=+L/2 are natural. So what you do is say that for an element of rod with a tiny length d\ell and charge \... 0 First I have to ask: does the question mention anything about the distance d? The reason why I am asking is because if d is large enough, we can say that it is in the far-field and we can easily approximate the field values using electrostatic theory treating the rod as point charge Q. I will provide an edit later if you want to use a far-field approximation.... 2 Here is another attempt at an answer, from a slightly different angle. It is less sophisticated and may add insight. You say that the capacitors are isolated before they are connected. Literally, that is only possible if they are in different universes. In the real world, they are connected through an infinitesimal capacitance - the capacitance between the ... 1 Charges at rest move when a force is applied on them and this is due to Newton's laws. Now to apply a force, we need a field, like electric/gravitational field. Each field acts upon certain measurable properties of a system, like gravitational on mass, electric on charge etc. Now potential is just a fancy name of height in electromagnetism. I hope you're ... 1 If you found the total x and y component of the total field just use: Magnitude E_{tot}:$$E_{tot}=sqrt(E^2_{tot-x}+E^2_{tot-y})$$The direction in angle:$$\theta = arctan(\frac{E_{tot-y}}{E_{tot-x}})$$The magnitude of force would just be$F=qE\$ Note: you should be careful in angles as it may sometimes not the angle measured from +x as the norm, for ...

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a) Add the x and y components separately, then calculate the magnitude E (the resultant field) and the angle it makes with the x axis. b) The force is qE, in the same direction as E.

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If the shell is conductive in this case, one thing for sure is that the net charge on the interior surface has the opposite sign and equal amount of charge as the charge placed inside of the cavity. If there are no net charge on the conductor before the charge inside of the cavity was placed, the charge on the conductor equal to the charge placed inside the ...

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Your statement is correct. The charge distribution is such that the hollow cavity of the conductor has a equal amount of negative charge induced on its inner part. This distribution is such that field due the cavity (including the charge inside the cavity) cancels out everywhere outside the cavity. So looking it the other way around external sources do not ...

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