Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
An electron in a P.P.capacitor is shot from the neg.plate to the pos. plate.
Why the book says that the negative plate is the lower potential? I thought that like charges repel thus, the electron would have a high potential when close to the neg.plate.
 one year ago
 one year ago
An electron in a P.P.capacitor is shot from the neg.plate to the pos. plate. Why the book says that the negative plate is the lower potential? I thought that like charges repel thus, the electron would have a high potential when close to the neg.plate.
 one year ago
 one year ago

This Question is Closed

vf321Best ResponseYou've already chosen the best response.1
No. Higher potential can be thought of as a hill and lower can be thought of as a valley FOR A POSITIVE CHARGE. Electrons have negative charge, so everything flips for them.
 one year ago

JamesWolfBest ResponseYou've already chosen the best response.0
So the book is wrong?
 one year ago

vf321Best ResponseYou've already chosen the best response.1
No. As I said, for an electron, the negative plate will have lower potential, since that means it will be a hill. The electron rolls off the hill to the valley, the positive plate.
 one year ago

vf321Best ResponseYou've already chosen the best response.1
** for an electron, the negative plate will be a hill since it has lower potential, not the other way around.
 one year ago

JamesWolfBest ResponseYou've already chosen the best response.0
I get you. So to escape the valley the electron would need kinetic energy exceeding the potential. An electron next to a negative plate needs less kinetic energy to escape than when its near a positive one.
 one year ago

vf321Best ResponseYou've already chosen the best response.1
Again, if a I have my negative plate at say x = 1 and my positive plate at x = 3, then voltage looks something like this: dw:1349756069232:dw Since \(\vec E =  \nabla V\), and, given the charge density \(\sigma\) on the left plate (and an equivalent opposite charge on the right) we can simplify it to \(E = \frac{\sigma}{\epsilon_0} = \frac{dV}{dx}\). So \(V=c (x1)\) where c is some positive constant. Put a proton anywhere on that line, and it will roll to the left. Put an electron on it, and you have to switch where it goes, since \(U=qV\), and with an electron q Is negative. Technically you have to plot \(U\), not \(V\) (which is independent of the charge in quesiton) if you want to look at kinetic energy and all that stuff about being able to roll from a valley, etc.
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.