calculate the electric dipole moment of an electron and proton 4.30mm apart?
99566
p=q*d\nq=1.60*10^-19 coulombs\nd=4.30*10^-3 m\nso\np = q*d = 1.60*10^-19 * 4.30*10^-3 = 6.89*10^-22 coulomb*meters
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.
27684
Protons, neutrons, and electrons.
660156
http://www.physlink.com/education/askexperts/ae25.cfm
558215
F = KM1M2/r^2\n\nwhere K is the constant, M1 and M2 are the magnetic moments, and r is the distance between them.
786060
First you need to look up the element on a periodic table of elements.  I am sure you have one laying around somewhere =P\nAnyway, then look in the square that contains your element for a number either in the upper left corner or in some cases directly above it.  That is the number of protons in the atom.  Then you also know the number of electrons which is exactly the same as the number of protons.  To figure out neutrons, look for another number on in the square that isnt in the left corner or above the element.  This is the atomic mass.  Take the atomic mass and subtract the number of protons to get the number of neutrons
393625
mass # =#protons + # neutrons\nand # protons = # electrons in an electronically neutral atom.\nso you can therefore find the # nuetrons\n#neutrons=mass#-# protons
241348
try:\n\n  \nFaraday constant  over Electron charge = Avogadro's number\n\n96,487 / 1.6022*10-19 = 6.022*10^23
247677
There's this equation called Coulomb's Law that relates charge and force: F=k*Q*Q/(r^2), where k is Coulomb's constant (8.988e9) and r is the distance between the two charges. That doesn't give us the units for an electric field, which is a force per charge. But we can divide both sides of Coulomb's Law by a Q, giving us F/q=k*Q/(r^2).\n\nTo find the magnitude and direction of the net electric field, you need to add together the electric field vectors created by each of the two charges. For the first charge, the electric field is k*15e-6/(.12^2)=9.36e6 N/C. The electric field of the second charge is k*10e-6/(.05^2)=3.60e7 N/C. Make sure to change cm to meters, and note that the charge q at which you are finding the electric field does not factor into our equation. Hopefully the problem describes in more detail the location of the two charges so you can determine the appropriate vector sum's magnitude and direction.
61553
well, the place to start is the periodic table.  uranium's configuration number is [Rn] 5f3 6d1 7s2, which means the outermost electrons are in the n = 7 shell.\nto calculate the radius of 7s electrons requires some seriously ugly quantum mechanics, so i'll just present you with some results:\nhttp://web.mit.edu/3.091/www/pt/pert1.html\nwhich states that the atomic radius is 1.38 angstroms.  other places on the web list 1.45 angstroms and 1.7 angstroms, so about 1.5 seems like a pretty good average.  (an angstrom is 10^-10 m, or 0.1 nanometer)\n\nas for the nuclear radius, this has to be measured from proton-scattering experiments.  typically they are around 1 femtometer, which is 10^-15m or 0.00001 of an angstrom.  (see the wikipedia page for more detail)\n\ncheers!
916756
non polar  \nit's linear in shape ,so the magnet moment equals zero \nthe lone pairs around the two O atoms \nevery O atom has 2 lone pairs\nthe central atom is C atom
719199
You are supposed to use symmetry and electronegativities to answer this. Draw the three dimensional structure and identify which atoms have the highest electronegativity. That will give you your dipole moment.\n\nIn this case, Br has a higher electronegativity than C and H, so the positive charge will be directed towards the C and the negative between the two Br atoms.
345042
Electric field:\n\nE=(1/4*pi*epsilon0)*(q/r^2)--> [spherical shell at r>Radius]\n\nE=0 --> [spherical shell at r<Radius]\n\nE=(1/4*pi*epsilon0)*(q/r^2) --> [spherical distriburion at r<=Radius]\n\nor E=(q/4*pi*e0*R^3)*r [uniform charge]\n\n-----------------------------------------------------\nPotential field:\n\nit doesn't have a formula but u can find it with this integral:\n\nv=(1/4*pi*e0)S(dq/r)-> S is integral!
561926
XeF  + e-  ============> Xe  + F-      E°/V = 3.4 with reference to the standard hydrogen electrode
977502
ok i fell asleep in physics, but...\n\n1/1837 th of the motion would be transferred to the atom.\n\nand 1836/1837ths would be the speed of the bounced electron.
179131
Try this:\nhttp://kr.cs.ait.ac.th/~radok/physics/f3.htm
945388
density,volume,weight,number of electron and proton.