A uniform electric field?
373192
f=m*a=E*q\na=Eq/m\n\nv^2=v0^2+2ax\n\nv=sqrt(v0^2+2ax)=sqrt(v0^2+2Eqx/m)\n\nv0=2.74e4\nE=2.58e3\nq=1.6e-19\nx=.00185\nm=1.67e-27
895113
This is true in isolated atoms.\n\nIf we put them into the electric fields generated by nearby atoms, especially those of water of crystallisation, the energy levels can change and rearrange the electron orbitals.\n\nTo explain this to our satisfaction would mean solving the equations and getting the answers.  Unfortunately this is known as a many bodied problem and only the simplest of them can be solved.\n\nTake heart in the fact that the energy levels are so close together that we can glibly announce the fact and move on quickly to the next topic!
360483
If it is a perfect conductor, it will have equal potential everywhere along its edge as charge will spread itself out evenly.
262366
Yes, but it is a small magnetic field
1013297
It boils down to the voltage drop across your body as a function of distance from the point of contact...\n\nAssume peak current of 30,000 amps and that the charge spreads out uniformly in all directions - as an expanding hemisphere.\n\nCurrent density will be 30,000/(2*pi*r^2)....roughly 2400/radius^2 in meters. (amps/meter^2)\n\nAt 100 meters that implies a current density of about .24amps/m^2.\n\nAssuming a conductance of 5mho/meter...that works out to a voltage drop of 1.2 volts per meter.\n\nYou're safe a football field away.  At 30 feet though, you might as well hold a paperclip in each hand and stick them in a wall-outlet.
2470
I'm assuming this is an electric dipole.\n\nSince a dipole can be modeled as a pair of opposite charges +_q separated by a small displacement, you can use the method of images and assume the field produced by the conductor is the same as the original dipole reflected across the plane with the signs of the charges reversed. That means the image dipole is in the same orientation as the original dipole and is at a distance r = 2I from it. The force is attractive (towards the plane). The energy between two dipoles having the same moment p parallel to their separation vector is:\n\nW = -2p^2/r^3\n\nwhere r is the separation distance. The force between two such real dipoles would be (i.e. this is the force on either one):\n\nF = dW/dr = -6p^2/r^4\n\nThis should be the same force experienced by the one dipole in the field produced by the conducting plane. Since r = 2I, this is:\n\nF = (3/8)p^2/I^4
414278
Any charged particle, like a proton or an electron is accelerated in an electrical field. One can make an electrical field by applying a large voltage differential between two electrical plates. Old cathode tubes televisions work like this: electrons are accelerated inside the tube and hit the screen. So chances are you have already an accelerator at home.\n\nAt CERN or Fermilab the particles are accelerated in mile wide rings. Therefore the encounter the same accelerating electrical field over and over. After millions of rounds they build up speeds close to the speed of light.
298380
the effect of electric fields on plants or other organic species
276990
read this page\nhttp://en.wikipedia.org/wiki/Field_(physics)
603730
The following refers to Coulomb force between two charges in a single uniform medium, not the two-medium situation described in  the question.\nAnswer 1 is helpful but the lack of exponent signs makes it misleading. The force is inversely proportional to the square of the distance, so the simplified formula is \nF = k * q1 * q2 / r^2 \nk is a constant = 8.99*10^9 N m^2 / C^2 (C is charge in Coulombs). Thus if q1 = q2 = 1 Coulomb, and q1 is separated from q2 by r = 1 meter, the force is 8.99*10^9 N or 2.01 billion pounds! \nThe above simplified formula is a contraction of the full formula (see ref.),\nF = q1 * q2 / (4 * pi * epsilon * r^2)\nin which k is replaced by 1/(4 * pi * epsilon), where epsilon is the product of epsilon0, the electrical permittivity of the vacuum (= 8.85 * 10^-12 c^2/(N m^2)), and kappa, the dielectric constant which = 1 for vacuum and is assumed in the simplified formula.\nFor two segments of distance as you specify, first solve for F1 and F2, each as a function of a particular distance (r or d - r) and dielectric constant. Then my inclination would be to try the sum-of-inverse square root solution which works for a similar situation, gravitational force F at distance R, computed from F1 and F2 at partial distances R1 and R2, where R = R1 + R2, \n1/SQRT(F) = 1/SQRT(F1) + 1/SQRT(F2)\nSorry I can't be more definite than that.
163366
Yes with a gaussmeter.
1037424
evenly distributed energy
343981
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.
355899
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.
459989
According to the Lorentz Force law:\n\nForce=q[E + v x B]\n\nso if there's no Electric field it will just be q(v x B) where q is the charge of the particle
241348
try:\n\n  \nFaraday constant  over Electron charge = Avogadro's number\n\n96,487 / 1.6022*10-19 = 6.022*10^23
504905
http://www.amasci.com/emotor/statelec.html\n\nhttp://www.phys.unsw.edu.au/PHYS1169/beilby/electric_fields.html\n\nhttp://users.pandora.be/educypedia/electronics/electrostatics.htm