In two charged plates, \nother than between the \nplates (and the edge\neffects), are there any \nfield line?
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In two charged plates, other than between the plates (and the edge effects), are there any field line? \nYES! other field lines are between the outer sides of the plates(depending on the charges of the plates and the convention used for the direction of the flow). To prove this, just connect the outer sides of the plates by a wire through an amper meter and you will see the meter moving.\n\nWhy is the electric field constant everywhere in the space between parallel plates? \nThey are not unless the area of both plates is inifinte and that the two plates are parallel.  In this case, the 1) direction of the field is the same every where since by definition the plates are parallel and 2) the absolute value is constant.  If not, there will be a surface current and the current would make the entire electric field (its magnitude) to be constant over the entire surface.
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There is no boundary as such because there is "nothing" beyond the edge; but the Universe in not infinite.  Difficult to get your head around I know especially as "nothing" in this case does not mean nothing as you understand it now.  If I go on you might blow a fuse so I will stop there.\n\n[re: above - "The Milky Way" is the name of our Galaxy not our Solar system]
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Sometimes. For instance, in circuit feeding a half-wave rectifier the impedance is much higher in one direction than the other.\n\nBut in general the impedances should closely match. IN fact, the electric system designed to have closely matched impedances.
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By symmetry, the electric field on the line through the center of the dipole is parallel to the dipole axis (the vertical components cancel out).  The field from -q is E_1 = -q/(4πe0r^2) and from +q is E_2 = -q/(4πe_0r^2).  The horizontal components of these fields are E_1cosø and E_2cosø where ø is the angle from each charge to the point, and r is the distance from each charge to the point.  If the dipole charges are a distance 2d apart (dipole moment of 2*q*d), the y coordinate of the point is √[r^2-d^2], and ø=arctan(y/d).  So r=√[y^2+d^2],and cosø= d/√[y^2+d^2.  Putting it all together you get\n\nE(y) = -[q/2πe_0] * 1/(y^2+d^2) * d/√[y^2+d^2] = -[q/(2πe_0)]*d*[x^2+d^2]^-1.5\n\nThe sign is negative because I assumed the + charge was on the positive x-axis, and the field is directed in the negative direction.
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An equipotential surface of an electrostatic field is given by:\n\nE = -∇Φ\n\nwhere φ(x,y,z) is the scalar field representing the electric potential at a given point. The symbol ∇represent the derivative of the scalar field Φ which results in a vector field that is called the gradient.\n\nSo for example, the electrostatic field of a point charge q is given by:\n\nE = (kq/r^2) ř\n\nwhere k is Coulomb's constant, and is = 1/4πε\nε is the Permittivity of free space\nand ř is the unit vector in the radial direction in polar coordinates\n\nHere we see that the equipotential surface of the electrostatic field of a point charge are "spheres" surrounding the charge q, and its strength decreases by 1/r^2. At some distance r from the charge q, the electrostatic field E has the same strength everywhere on the surface of the sphere.
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It's either just one charge: The electrical, which appears in positive and negative characteristics, or if you see "charge" in a more general way it's more then one charge: magnetism has also a source, but that appears always in dipoles, mass is the "charge" of gravity, but that appears just with a positive sign, and the strong interaction has three kind of charges, called "colour charges": red green and blue.\n\nTwo see an underlaying principle for this "chaos" we need the "Grand Unified Theory" which still is "work in progress" ;)
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The incident ray is perpendicular to the surface of reflection.\nThen one has the least distance case.
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Intensity varies as the square of the distance.\nThat should help
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Let the two +Q charges be placed left to right with the distance r separating them.  Let a third charge of unknown charge q be placed a distance x to the right of the first charge.  Now, look at each of the +Q charges and sum the electric force acting on them.  For example, the left hand +Q charge experiences a repellent force from the +Q on the right, pushing it towards the left.  To balance this out, there must be an attractive force pulling it to the right, so the q charge is negative.  However, this is just an observation.  In order to maintain the integrity of the algebra, we should continue to assume that q is positive, but recall the negative result llater when we check our answer..  The proportionality factor for electric field strength can be ignored, because it will cancel out, so you can treat Q/r^2 as the value under consideration.  In this case, we need Q/r^2 + q/x^2 = 0.  However, because the +Q charges are equal in magnitude, I will make the assumption that the distance x = r/2, so Q/r^2 = -4q/r^2 ==> Q = -4q ==> q = -Q/4.  Perform the same force balance on the right hand +Q charge to verify.\n\nIf the initial charges had been unequal, x would have remained a variable, and the complementary distance (r - x) would have been used for the second force balance in order to form a system of equations.
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Assuming the plates are also superconductors and the dielectric between them is vacuum, the only place for the energy to go is by electromagnetic radiation. The connecting superconducting wire will have inductance so you will have a resonant "tank" circuit with a very high "Q" factor. What you get is damped oscillations at the resonant frequency. Depending on dimensions, it could take a while for all the energy to radiate away, but radiate it will.
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electrostatics, charge need not be 'static' in the sense of unchanging. Instead 'static' implies that the dynamic coupling between electric and magnetic fields can be ignored. In electrostatics we study e-fields, voltage, and charge, but ignore any magnetic fields generated by the motion of these charges or that may be present for other reasons. Because of the electric field's relationship to and interaction with magnetism, electrostatics is a subfield of electromagnetism.\nThe electrostatic approximation\nThe validity of the electrostatic approximation rests on the assumption that the electric field is irrotational:\n\n \n\nFrom Faraday's law, this assumption implies the absence or near-absence of time-varying magnetic fields:\n\n \n\nIn other words, electrostatics does not require the absence of magnetic fields or electric currents. Rather, if magnetic fields or electric currents do exist, they must not change with time, or in the worst-case, they must change with time only very slowly.\n\nIn some problems, both electrostatics and magnetostatics may be required for accuratete predictions, but the coupling between the two can still be ignored.\n\n[edit]\nElectrostatic potential\nBecause the electric field is irrotational,is possible to express the electric field as the gradient of a scalar function, called the electrostatic potential (also known as the voltage). Thus, the electrostatic potential Φ is related to the electric field E by the equation:\n\n \n[edit]\nFundamental concepts\n[edit]\nCoulomb's law\nThe fundamental equation of electrostatics is Coulomb's law, which describes the force between two point charges:\n\n \n[edit]\nThe electric field\nThe electric field (in units of volts per meter) is defined as the force (in newtons) per unit charge (in coulombs). From this definition and Coulomb's law, it follows that the magnitude of the electric field E created by a single point charge Q is:\n\n \n[edit]\nGauss's law\nGauss' law states that "the total electric flux through a closed surface is proportional to the total electric charge enclosed within the surface." The constant of proportionality is the permittivity of free space.\n\nMathematically, Gauss's law takes the form of an integral equation:\n\n \n\nAlternatively, in differential form, the equation becomes\n\n \n[edit]\nPoisson's equation\nThe definition of electrostatic potential, combined with the differential form of Gauss's law (above), provides a relationship between the potential Φ and the charge density ρ:\n\n \n\nThis relationship is a form of Poisson's equation.\n\n[edit]\nLaplace's equation\nIn the absence of unpaired electric charge, the equation becomes\n\n \n\nwhich is Laplace's equation.\n\n[edit]\nStatic charge generation\n[edit]\nCharge separation by contact\nThe presence of surface charge imbalance means that the objects will exhibit attractive or repulsive forces. This surface charge imbalance, which leads to static electricity, can be generated by touching two differing surfaces together and then separating them due to the phenomena of contact electrification and the triboelectric effect. Rubbing two non-conductive objects generates a great amount of static electricity. This is not just the result of friction; two non-conductive surfaces can become charged by just being placed one on top of the other. Since most surfaces have a rough texture, it takes longer to achieve charging through contact than through rubbing. Rubbing objects together increases amount of adhesive contact between the two surfaces. Usually insulators, i.e., substances that do not conduct electricity, are good at both generating, and holding, a surface charge. Some examples of these substances are rubber, plastic, glass, and pith. Conductive objects only rarely generate charge imbalance except, for example, when a metal surface is impacted by solid or liquid nonconductors. The charge that is trans
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In a linear medium you can work out:\nThe force between X and Z  charges  using Coulomb's Law\nThe force between Y and Z  charges using Coulomb's Law\n\nThen add the Forces vectorally taking into account the directions.\n\nThe Question doesn't mention the geometry (angles) of the set up in which case I would imagine you just add them algebraically.
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It depends on how far away from the noble gases they are. Like anything in gourp IA would be 1+, IIA 2+ etc, then when theyre closer to the right then its like VIIA 1-, VIA 2-\n\nTransition metals are different because they form multiple ions. It would indicate what the charge is.
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The first one isn't, because it's undefined when x=4 (assuming the (x-4) is the denominator, not just the x)\n\nThe second one is; it's a horizontal line with no breaks or holes.
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The electric field at a point in space is equal to force per unit charge exerted on a positive test charge placed at that point.\n\nConsider a point charge +q located at point. The charge +q sets up electric field in the space surrounding\nit. If a small positive test charge +Q placed at point P experiences a force F.Then electric field ,E=F/Q\nMind the following:\n(i) Electric field (force/charge) is a vector quantity, i.e., it has magnitude as well as direction.\n(ii) The SI unit of electric field is newtons per coulomb (N/C). As shown later, it can also be expressed\nas volts per metre (V/m).\n(iii)The other terminology used for electric field is electric field intensity or electric field strength.
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no, by definition equipotential lines will always run parallel to each other. That is when they are close together. So they can never cross.
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The movement of an electron from one place to another.