Dude, my teacher said the exact same thing, but I wasn't paying attention, because seriously, that guy sucks. But when you said it, I paid attention, and now it makes sense. Thanks bud. I owe you one.
Thanks for the wonderful explanation! I got an identical question to this, except with part B, which asked to calculate the velocity as the electron leaves the parallel plates. In the solution, to find the y-component of velocity they used vfy= ay*t, while I did vfy = dy/t. Why is that wrong? Your kind explanation would be much appreciated. Many thanks
The y velocity isn't constant, so you can't just say v=distance/time, you have to use the constant acceleration kinematics approach. I assume you got dy by doing this, but then the thing you computed was an *average* y velocity instead of the final y velocity. z
Side view, but the force of gravity is less than the electric force by 14 orders of magnitude. Calculate mg vs. qE to get an estimate. I should have mentioned that, but it's the kind of thing you start to take for granted with the strength of electric forces vs. gravitational! -- Zak
I don't really know how to get mathematical with it, but if the electron hits the + plate, I think it would just slightly reduce the potential difference between the plates, then the power source would pump one more electron across the plates to get back to the original potential difference. Multiply this effect by thousands, and the power source must constantly work to maintain the potential difference. I haven't done a bunch of research on this sort of experiment, but that's my guess! z
Good question! The electric field is doing positive work on the electron (potential energy is decreasing) as it moves from lower potential to higher potential (closer to the positive plate). Alternatively, arguing it with forces, you could say the component of the electric force parallel to the displacement does positive work for any increment of time. As the path bends, you end up with a component of electric force in the direction of displacement at each increment, and this contributes positive work. Another way to gain insight here: this is really just projectile motion with a horizontal launch (upside down due to the direction of electric force). If you launch a projectile horizontally from the top of a building, it gains kinetic energy as the path bends, and that's from the drop in gravitational PE, or if you prefer the force/work analysis, because a component of the force of gravity points in the direction of displacement for each time increment. -- Zak
@@ZaksLab Thanks for your answer! I still do not really get it. If the EM-field is doing positive work on the electron, this means electric potential energy is converted into kinetic energy (potential energy of the capacitor, I guess?). But nothing in the capacitor change. The electric potential energy in the capacitor stays the same. So my problem is this: The electron has increased its kinetic energy by being deflected by the capacitor, but the capacitor itself has not lost any energy. This would mean I could let an infinite number of electrons pass the capacitor, all increasing their kinetic energy, without any energy bill to pay! This looks very much like a clear violation of the energy principle. If you look at the gravitational analogy. A stone accelerating to the ground increase its kinetic energy, at the expense of the gravitational potential energy of the entire system. If you look at the deflection analogy. A spacecraft can increase its kinetic energy by the slingshot effect. This can only be done at the expense of the planet losing kinetic energy. Where is the fault in my reasoning??? Best regards! Bengt-Erik
@@bengt-erikandersson2930 I think the answer lies in the approximations we get used to making with capacitors; i.e., they are "infinite plane" capacitors. In reality the fields fringe at the ends and the equipotential lines loop around. If you pass electrons through a finite capacitor, they will still speed up (which is fair: negative charges getting farther from other negatives/closer to positives *should* speed up in the same way masses falling to the ground do), but the electrons will start crossing back over equipotential lines as they get far from the capacitor, and that will slow them back down. In the "small capacitor" limit, the electron is approaching a dipole from a great distance, briefly speeding up as it gets closer to the + side, then taking off to infinity again where it will slow back down to the original speed. These are good fundamental questions, and it would be cool to work out all the details of the speed of an electron passing through a finite capacitor by looking at electric potential energy at each instant in time. That problem is above my pay grade, but it's something that might be well suited to a coding lab to create the simulation/animation.
@@ZaksLab I think you are on to something there (fringe effects). I always had problems with these sharp edges of the electric field. Sharp edges newer appear in nature. I thank you for your answers it was very helpful.
Dude, my teacher said the exact same thing, but I wasn't paying attention, because seriously, that guy sucks. But when you said it, I paid attention, and now it makes sense. Thanks bud. I owe you one.
you're welcome! -- Zak
Oh man, this has got to be one of my favorite movies ever. The plot twist gets me every time! Thanks so much for creating this masterpiece!!
next time relativistic speed is the twist! thanks z
You just saved my my physics grade you absolute legend
haha thanks! z
Thank you my boy due to solve this valuable problem.
You bloody legend:D
Haha . . . in my own mind. Glad you liked it though! -z
Thanks for the wonderful explanation! I got an identical question to this, except with part B, which asked to calculate the velocity as the electron leaves the parallel plates. In the solution, to find the y-component of velocity they used vfy= ay*t, while I did vfy = dy/t. Why is that wrong? Your kind explanation would be much appreciated. Many thanks
The y velocity isn't constant, so you can't just say v=distance/time, you have to use the constant acceleration kinematics approach. I assume you got dy by doing this, but then the thing you computed was an *average* y velocity instead of the final y velocity. z
Thank you so much - I now understand why.
is the diagram a top view or a side view - bc if it's a side view then wouldnt we have to take into consideration gravity?
Side view, but the force of gravity is less than the electric force by 14 orders of magnitude. Calculate mg vs. qE to get an estimate. I should have mentioned that, but it's the kind of thing you start to take for granted with the strength of electric forces vs. gravitational! -- Zak
@@ZaksLab lool truee thank you man ^_^
is that possible that electron fall on the plate in its journey? (with mathematically approach, please). thanks
I don't really know how to get mathematical with it, but if the electron hits the + plate, I think it would just slightly reduce the potential difference between the plates, then the power source would pump one more electron across the plates to get back to the original potential difference. Multiply this effect by thousands, and the power source must constantly work to maintain the potential difference. I haven't done a bunch of research on this sort of experiment, but that's my guess! z
@@ZaksLab thanks a lots zak.
thank a lots. Zak.
The electron exit the plates with an increas in kinetic energy. Where does this energy comes from?
Good question! The electric field is doing positive work on the electron (potential energy is decreasing) as it moves from lower potential to higher potential (closer to the positive plate). Alternatively, arguing it with forces, you could say the component of the electric force parallel to the displacement does positive work for any increment of time. As the path bends, you end up with a component of electric force in the direction of displacement at each increment, and this contributes positive work.
Another way to gain insight here: this is really just projectile motion with a horizontal launch (upside down due to the direction of electric force). If you launch a projectile horizontally from the top of a building, it gains kinetic energy as the path bends, and that's from the drop in gravitational PE, or if you prefer the force/work analysis, because a component of the force of gravity points in the direction of displacement for each time increment.
-- Zak
@@ZaksLab Thanks for your answer!
I still do not really get it. If the EM-field is doing positive work on the electron, this means electric potential energy is converted into kinetic energy (potential energy of the capacitor, I guess?). But nothing in the capacitor change. The electric potential energy in the capacitor stays the same. So my problem is this: The electron has increased its kinetic energy by being deflected by the capacitor, but the capacitor itself has not lost any energy. This would mean I could let an infinite number of electrons pass the capacitor, all increasing their kinetic energy, without any energy bill to pay!
This looks very much like a clear violation of the energy principle.
If you look at the gravitational analogy. A stone accelerating to the ground increase its kinetic energy, at the expense of the gravitational potential energy of the entire system. If you look at the deflection analogy. A spacecraft can increase its kinetic energy by the slingshot effect. This can only be done at the expense of the planet losing kinetic energy.
Where is the fault in my reasoning???
Best regards!
Bengt-Erik
@@bengt-erikandersson2930 I think the answer lies in the approximations we get used to making with capacitors; i.e., they are "infinite plane" capacitors. In reality the fields fringe at the ends and the equipotential lines loop around. If you pass electrons through a finite capacitor, they will still speed up (which is fair: negative charges getting farther from other negatives/closer to positives *should* speed up in the same way masses falling to the ground do), but the electrons will start crossing back over equipotential lines as they get far from the capacitor, and that will slow them back down. In the "small capacitor" limit, the electron is approaching a dipole from a great distance, briefly speeding up as it gets closer to the + side, then taking off to infinity again where it will slow back down to the original speed. These are good fundamental questions, and it would be cool to work out all the details of the speed of an electron passing through a finite capacitor by looking at electric potential energy at each instant in time. That problem is above my pay grade, but it's something that might be well suited to a coding lab to create the simulation/animation.
@@ZaksLab I think you are on to something there (fringe effects). I always had problems with these sharp edges of the electric field. Sharp edges newer appear in nature. I thank you for your answers it was very helpful.