Inf. Parallel Plate Capacitors and Potential Energy

AI Thread Summary
The discussion focuses on calculating the kinetic and potential energy of an electron moving between two infinite parallel plates separated by 5.2 cm with a potential difference of 150 V. The kinetic energy at plate X can be determined using the formula U = qV, where the potential energy at plate W is converted into kinetic energy as the electron accelerates. Once the electron exits the plates, the electric field is zero, indicating that its kinetic energy remains constant beyond this point. The participants also discuss the principles of superposition and electric field calculations, emphasizing the need for understanding the charge density of the plates to fully solve the problem. Overall, the conversation highlights key concepts in electrostatics and energy conservation in the context of parallel plate capacitors.
kevtimc25
Messages
5
Reaction score
0
Hopefully no one minds that I post a couple questions in a row as long as I'm following the rules:

1. Homework Statement


phyd.png

Two infinite parallel plates labelled W and X are separated by 5.2 cm. The electric potential between the plates is 150 V. An electron starts from rest at time tW and reaches plate X at time tX. The electron continues through the opening and reaches point P at time tP

c.) Find the electron’s kinetic energy and potential energy at point *P.
d.) Find the minimum speed of the electron at time tw needed for it to escape to infinity.


Homework Equations


U = qV , energy conserved

The Attempt at a Solution


We can calc. the KE X easily: Ux = 0, Ex = Uw, Uw = 150V * -1.60*10^-19 C = 1/2 m(e) v^2 so we have v and there is no electric field once it is outside of the plates. A previous given answer indicates that the speed is constant once we hit X (I don't understand why other than that no outside forces are acting on the electron if we exclude gravity), so how can we calculate U and K at *P?
 
Physics news on Phys.org
Do we know the initial position of the electron?
If the electron is accelerated through a potential difference of V then it's kinetic energy is V electron-volts.

Per your question: do you know how to calculate the potential due to a single plate?
Do you know the superposition principle?
 
Yes it is at rest at W.

I know the EF is σ/2e0 everywhere due to each plate and you can find V by integrating E through a distance. And I know you can also find V by adding the potentials from each plate. But we don't now the Surface density of the plates.
 
You also know the potential difference between the plates is 150V... this gives you the kinetic energy at X.
Per the charge density - just put it equal to ##+\sigma## for the +ve plate and ##-\sigma## for the -ve plate.
 
Right, outside of the PPC the EF is 0 everywhere, so the ΔV would have to be 0 right? That means the potential doesn't change so the KE can't change as well (at point X and beyond).
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top