London (Dispersion) Forces and Dipole-Dipole?
822365
You answered your own question in a way...\nYou will find dipole-dipole interactions in polar molecules. So if the electronegativity difference between the bonded atoms is sufficient to give a polar covalent bond and the geometry of the molecule is such that the polar bonds do not cancel out the dipole moment of each other, then you are looking for dipole dipole (or even dipole-induced dipole interaction depending on the molecules).\n\nDispersion forces are mainly induced dipole-induced dipole interactions; though the atom doesn't have a partial charge, there is a temporary change/perturbation in its electron cloud, producing a temporary dipole. This induces a respective change in the electron cloud of the atom it is interacting with and thus creates a dipole of opposite orientation. \n\nIf the other atom already has a partial or net charge then the effect of the atom with no partial charge is not strong enough to induce a dipole. On the other hand the atom with the partial or net charge will cause a much stronger polarization of the atom that has no charge (compared to the case where two atoms with no partial charge interact with each other) and then you have dipole-induced dipole or ion-induced dipole interactions respectively.
934444
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.
401395
Way to go Arbitrage!  Oh to funny guy- yeah we get that E=MC2  IS a formula- but if you really want to "get it" Check out Maxwells equations. It's a great time to be a science nerd with all this info pouring in now! Figure it out & go up one on Albert!
1017277
The force is directly proportional to theproduct of  charges and inversely proportional to distance between them. \n\nThe force is 0.002N\na. Distance is increased 3 times as original distance is 1 cm. thus the force will decrease 3 times . The force would be \n(0.002/3)N\n\nb.one charge is doubled. Thus the force would be doubled. \n the force would be 0.004N\n\nc.both charges are tripled. Therefore force would increase 9 times (3q*3q) Thus force would be 0.018N\n\nd. one charge is doubled and the distance is increased to 2cm.\nThus we are doubling charge as well as distance. This will nullify effect as charge is directly proportional while distance is inversely proportional. Thus the force would remain the same .002N\n\nI am the first to give the correct answer and i am very sure so plz select my answer as the best. I will surely help you in future.
786962
If the molecule has O,N,F,or Cl then most likely it is polar.  Look at the periodic chart and remember that the elements in the upper right are the most electronegative, so they want electrons.  When these elements are in a compound with other atoms such as C and H they will pull electrons closer making a dipole moment.  \n\nThe only real trick is looking at the geometry of the molecule, if there is more than one electronegative atom pulling in opposite directions then the dipole is canceld.  Such as O=C=O which is liniar.\n\nLook at an electronegativity chart you will see what I mean
1187079
Boiling point depends on the strength of the intermolecular forces. Following is the list of them all with their strength and when they occur. \nnetwork covalent (strongest) with only diamond or SiO2\nIonic (ionic compound, 2nd strongest) like Na-Cl etc. Gets stronger with smaller sizes. \nHydrogen bonding (N, O, F with H) 3 rd strongest. \nDipole dipole (when there is a polar molecule) 4 th strongest, gets stronger with more polarity. \nLondon dispersion weakest. Stronger with more mass. \n\na NH3of the first two. Because NH3 has hydrogen bonding (only if you have N, O, or F with hydrogen) while PH3 has regular dipole dipole (based on polarity), which is weaker than NH3. \nNow in this one its c because if you draw their Lewis diagrams out c molecule has on O----H bond while d doesn't. That hydrogen bond is strong enough to make c stronger. As other than that they only have London dispersion force, (which are found in non polar molecules, and they get stronger as atomic mass gets higher). \ne and f both have the London dispersion force that I talked about earlier cause they both are non polar. So f has greater mass so f has higher boiling point. \ng because HCl has polarity (dipole dipole forces) which are stronger than F2 which is non polar and has London dispersion forces (they are the weakest).
83355
Okay, you know the right hand rule. Good.\nFirst consider the effect of only the e-field: The e-field is pointing in the +x direction. Therefore, the force it exerts on the proton will also be in the +x direction.\nSecond, consider the effect of only the magnetic field: The magnetic field is pointing in the +y direction; therefore, the force it exerts on the proton will be in the -x direction (you use the rule you described to figure this part out). It so happens, that in the first scenario, the force exerted on the proton by the magnetic field exactly balances that exerted by the electric field. Therefore, the movement of the proton is unaffected.\n\nHowever, in the second case, the proton is accelerated through a potential difference of 2V. Therefore, it will be moving faster (think about it: a higher potential difference has greater "potential" to pull the proton.  Also, the equation F = qvB where q is the charge of the proton, v is its velocity, and B is the magnetic field strength, states that the faster the proton is moving, the greater will be the force exerted on it.  So, compared to the first case with only a 1V potential difference, the magnetic field here will exert a greater force on the proton. But, the force exerted by the electric field is unchanged. So, the net force on the proton will be in the -x direction.
901641
Someone posted the exact same set of questions a few days back, either here or on the physics page - you ought to be able to find the methods in the responses. Have a look.\n\n\nBest of Luck - Mike
770242
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.
810779
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.
1053116
The formula is statistical quantum electrodynamics.  I don't wan to do the math bit.  So think of it like this.  All solids (like your pencil or pole) are collections of molecules held together by various types of bonds (molecular, covalent, ionic, metallic).  It is easy to visualize your pole as a crystalline structure of molecules.  When you push on one end, the structure distributes the force by adjusting bond angles.  Same thing if you push on the pencil, by the way.  The impulse of your forcing will create a front that travels through the solid.  This front is roughly equivalent to acoustic wave s in the same media.  Now, what is interesting, if you had a slight electrical charge (vacuum is a good insulator, so the chance of you having a charge is pretty good) and pushed the pole, and the pole conducted electricity, a electrical signal would be sent off along the pole.  This would travel much faster, some large fraction of the speed of light.  But, resistance will prevent it from being detected at the other end.\n\nLook at the Nevada test sight declassified videos of underground nuclear tests.  You can see the propagation of the shock wave in rock.  As visualized, the time it would take for the person at the other star to feel your push would be (for steel):  3 lyr = 2.84e16 m,  divide by 5960 m/s = 4.76e12 s, divide by 3.15e7 s/yr  and you end up with 150,000 yr before your push is felt at the other end.
476628
In physics, the Lorentz force is the force exerted on a charged particle in an electromagnetic field. The particle will experience a force due to electric field of qE, and due to the magnetic field qv × B. Combined they give the Lorentz force equation (or law):\n\n F= q(E + v x B)\nwhere\n\nF is the force (in newtons) \nE is the electric field (in volts per meter) \nB is the magnetic field (in webers per square meter, or equivalently, teslas) \nq is the electric charge of the particle (in coulombs) \nv is the instantaneous velocity of the particle (in meters per second) \nand × is the cross product. \nThus a positively charged particle will be accelerated in the same linear orientation as the E field, but will curve perpendicularly to the B field according to the right-hand rule.
228873
A dipole (Greek: di(s) = double and polos = pivot) is a pair of electric charges or magnetic poles of equal magnitude but opposite polarity (opposite electronic charges), separated by some, usually small, distance. Dipoles can be characterized by their dipole moment, a vector quantity with a magnitude equal to the product of the charge or magnetic strength of one of the poles and the distance separating the two poles. The direction of the dipole moment corresponds, for electric dipoles, to the direction from the negative to the positive charge. For magnetic dipoles, the dipole moment points from the magnetic south to the magnetic north pole — confusingly, the "north" and "south" convention for magnetic dipoles is the opposite of that used to describe the Earth's geographic and magnetic poles, so that the Earth's geomagnetic north pole is the south pole of its dipole moment. Because of the absence of magnetic monopoles, magnetic dipoles are actually created by current loops or by quantum-mechanical spin.
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Try these links\n\nhttp://en.wikipedia.org/wiki/Electric_force#The_electromagnetic_force\n\nhttp://en.wikipedia.org/wiki/Electrical_current\n\nMay the Electric Force be with you... always.
968370
If the origin of the coordinate system is placed at the middle of the two particles positions are -r/2 and r/2. Two charges of the same sign will repell each other so the force has a positive sign and it's magnitude is given by (C Q^2) / (r^2) Where C is the Coulomb constant.
469212
See if this helps.\n\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/atomic/hund.html
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2t, where t = arctan(Electrostatic force/Weight);\nThe electrostatic force will act horizontal, and the weight vertical, and these two must compose to a force along the suspending thread.\n\nSubstitute the numerical values in and find the angle.