Urgent! Great question in electricity deflection of proton-you wanna answer it?
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.
508002
electricty is form of energy which is created by flow of positive charges(conventional assumption).but actually it is due to the flow ofelectrons inthe direction opposite to the flow of electricity.actually electrons do not travel with that speeed the electric field does the necessary.\n\nbut u have mentioned that matter particles cannot atttain the velocity comparable to that of light.\ni think its not correct\naccording to einstien's theory of relativity\nno particle in this universe is able to attain the velocity greater than that of light even same velocity for some.\nbut if the particle travels with a velocity comparable to that of light then the mass changes accordingly\nm=mo/sqrt(1-v^2/c^2)\nwhere mo is the rest mass of particle\nv is velocity of particle\nc is the velocity of light\nm is the mass of particle at velocity v
779247
no... protons are held in place by the negatively charged electrons.\nits called neutron radiation or ionizing radiation. This type of radiation is one of the worse because it can pass through matter better than other types and only stops after a headon collision with a nucleus\n\nDiscovery\n\nIn 1930 Walther Bothe and H. Becker in Germany found that if the very energetic alpha particles emitted from polonium fell on certain of the light elements, specifically beryllium, boron, or lithium, an unusually penetrating radiation was produced. At first this radiation was thought to be gamma radiation although it was more penetrating than any gamma rays known, and the details of experimental results were very difficult to interpret on this basis. The next important contribution was reported in 1932 by Irène Joliot-Curie and Frédéric Joliot in Paris. They showed that if this unknown radiation fell on paraffin or any other hydrogen-containing compound it ejected protons of very high energy. This was not in itself inconsistent with the assumed gamma ray nature of the new radiation, but detailed quantitative analysis of the data became increasingly difficult to reconcile with such a hypothesis. Finally (later in 1932) the physicist James Chadwick in England performed a series of experiments showing that the gamma ray hypothesis was untenable. He suggested that in fact the new radiation consisted of uncharged particles of approximately the mass of the proton, and he performed a series of experiments verifying his suggestion. Such uncharged particles were eventually called neutrons, apparently from the Latin root for neutral and the Greek ending -on (by imitation of electron and proton).
199003
a proton has a weight of approximately 1836.15 electrons
854263
The mass of the electron has been measured to an accuracy of 0.17 parts per million.  That's consistent with all electrons having the same mass, but not enough to rule out the possibility of more than one mass value where all are within the accuracy limits of our measurements.  The proton mass is known to the same accuracy.\n\nThe charge of the electron has been measured to 85 parts per billion.  No measurement has found a difference in magnitude between the electron and proton charge.\n\nTo answer the question, I would try to think of experiments whose outcome would be different if the mass or charge differed.  I would also run lots of simulations with varying assumptions for those values.  Quantum effects make it extremely difficult to measure the results of such experiments.
1086480
Question:  Exactly how fast do electrons travel?\n------------------------------------------------\nAnswer:  Electrons can have a wide range of speeds.  A slow case: we know that\nelectrons move when there is a current flow in a wire, but the speed at\nwhich the electrons themselves move in the wire -- the so-called electron\ndrift velocity -- surprises most people.  For example, for a copper wire of\nradius 1 mm carrying a steady current of 10 amps, the drift velocity is\nonly about 0.024 cm/sec ! On the fast side: the Bohr model of the hydrogen\natom has the (bound) electron zipping around the nucleus at about 2 million\nmeters/sec.  And on the very fast side, some examples are: beta particles,\nwhich are emitted by some radioactive materials; and the innermost elec-\ntrons of atoms of elements having large atomic number, such as uranium.  In\nthese cases the electrons are traveling at very nearly the speed of light.\n(about 300 million meters/sec).\nRcwinther
1141746
Electric charge, linear momentum and angular momentum are all absolutely conserved - meaning all of the time.  The problem is that neither number of protons nor the number of protons and neutrons must necessarily be conserved in a reaction.  However, the number of baryons (essentially the number of "large" particles, in this case protons and neutrons) is not conserved only in certain specific reactions.  My guess is that your teacher either doesn't know that or is simplifying things for you.  Consequently, I'd go with "number of protons".
962112
Yes the electricity is electron flow,which is the  number of electrons flowing in a cross section of a conductor for one oscilation .\nElectrons have an electrical mass which is equal to \n9.1 x10^-31kg.\nso if you take the number of electron per oscilation times the mass of the electron you have the total electrical mass flowing,\nYou take the total mass(assuming acceleration relative to the earth's center of mass)times 9.8 m/sec^2 would equal the weight of the electrical mass flowing.Hence a used battery would weight less.\nFor the simple reason that in the reaction gas has formed and evaporated and the excess electrons have dissipated and lost mass in the form of heat.\nThe weight difference can only be calculated. since the mass loss is very minute.The mass lost would be energy expanded by the battery divided by C^2.
15507
Depends upon the experimental conditions.  Electrons are not hard little balls with a negative charge.  They are best thought of as areas of possibility for the location of a wave.  The angle of interaction, the relative velocity of the electron and proton and many other factors will determine the observed effect of their interaction.  Some believe that the expectations of the observer also plays a big role.  Neutrons have been “observed” to breakdown into an electron and a proton under some conditions.  The reverse seems possible.
469436
READ PATEINTLY.. IT MAY APPEAR LONG BUT I GURANTEE YOU SATISFACTION....\n\nyes it is safe....\n\n\nBUT NOT BECAUSE OF RUBBER TYRES..... REMEMBER...\n    \n    NOT BECAUSE OF RUBBER TYRES...\n\nActually when lightening strikes earth surface the charge is in search of the shortest path (least resistance path) to reach the zero potential surface that is earth. Now a car with metallic body is a very good thing to provide that short route to the ground. Same concept as that of a "lighting rod" placed at the top of high rise buildings which attract the lightning towards it instead of the building concrete.\n\nI assume that you know that A PERSON WILL NOT FEEL ANY SHOCK UNTILL AND UNLESS THERE IS FLOW OF CURRENT THROUGH HIS BODY PARTS...\n\nWhen light strikes on the car top, it will get dispersed in all directions and flow through the frame to the wheel frames and then jump from there to the road through the wet tyres. This short jump (of around 8 cm height), in any case, requires less work than conducting electricity through highly non conductive air for height of about a car (120 cm).. or pass through you head and get out from your shoes...\n\nNow imagine yourself sitting in car... you are not in contact from the top... so when light strikes it has 2 options... either to jump  to your head or flow through the conducting car body... answer is obvious and it will pass through the car body...\n\nNow if my legs are touching the car floor then why not i get a shock then..? so answer again is that now when charge is at the floor it is having a zero potential surface below it and there is no point entering into your body and move unneccesarily upwards...\n\nHope you got it...
536354
Electric field, E, is\n\nE = K*q/(r^2)\n\nAt the "neutral point" the electric fields from each point will be equal in magnitude, but opposite in direction, and sum to zero (a charged particle placed at this point would be equally attracted to each point charge).  That is:\n\nE1 + E2 = 0\nE1 = -E2\n\nThe form E is give in above is a scalor, not a vector, so has no direction- therefore, we'll say that the magnitude of |E1| = |E2|\n\nK*Q/(r1^2) = K*4Q/(r2^2)   \n1/(r1^2) = 4/(r2^2)       (eq 1)\n\nI'll assume that this is linear distance, since I don't have any other information. This means that \n\nr1+ r2 = 12 cm             (eq 2)\n\nSolving equation 1 for r2 gives:\n((r1^2)*4)^0.5 = (r2^2)^0.5  \n\nr2 = r1*2\n\nsubstituting into equation 2:\n r1 +r1*2 = 12cm\n\nr1 = 4cm\n\nSo the distance from Q to the neutral point is 4 cm
672475
Since the proton is moving in a circular pattern, the force given is the centripetal force.  I know that it it worded weirdly, but I believe that is just to throw you off.  The equation for the centripetal force due to circular motion is F = m * (v^2) / r  \n\nwhere F is the force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.\n\nSo we set up the following:\n\n8.00 * 10^-13 N = (1.67 * 10 ^-27 kg) * (v^2) / (0.8 m) \n\nI used 0.8m instead of 80cm because we need to be in standard SI units.  This will make sense later. Rearrange and solve for v.\n\nV = sqrt[ (8.00 * 10^-13 N) * (0.8 m) / (1.67 * 10^-27 kg) ]\n\nSince N = (kg*m)/(s^2)\n\nwe can re-write as:\n\nV = sqrt[ (8.00 * 10^-13 (kg*m)/(s^2)) * (0.8 m) / (1.67 * 10^-27 kg) ]\n\nthe kg cancel out:\n\nV = sqrt[ (8.00 * 10^-13 (m)/(s^2)) * (0.8 m) / (1.67 * 10^-27) ]\n\nSimplifying you get:\n\nV = sqrt(3.83 * 10^-14 m^2/s^2)\n\nSince they tell you that the sqareroot of 3.83 is 1.96, then we get our final answer to be:\n\nV = 1.96*10^7 m/s\n\nLet me know if you need furhter clarification on this problem.
899593
Hello,\n\nIn this link you will have what you need:\n\nhttp://www.bsu.edu/web/fseec/pie/Lessons2/Physics/Potato%20Battery.doc\n\nI believe now will be OK.
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.
1151404
If the deflection surface has any diffraction or diffusion to it, the light may just be redirected imperfectly. 12kV? Jumpin Gigawatts!
866700
electricity produced due to flow of charges/electrons
988797
Are you sure you are not mixing up your neutrino indexes? I believe that the neutrons are the ones that give off the more energy at rest, as electrons pass by they convey energy to other protons which is how elements are able to hold and give charges (positively and negatively charged ions). When a proton can pass energy to a neutron, the electrical balance is changed (+/-). Electons are then displaced, and (if the components involved are able to rearrange the atoms to create a different element structure, while giving off electrically charged ions), a fusion of chemical and electrical energy takes place releasing energy at a higher at-rest-rate (although this has not yet been scientifically proven, as this requires elements to approach the speed of light - leading to a comparitive slowdown of particles). This then produces photons (as such that are produced by UV/ X rays, and exist within light) which are more of a product of the fusion process, as opposed to the neutrino reaction within the core of atomic fusion.\n\nHope this helps with the confusion.