Can Electrons Leave a Circuit Through an Open Branch?

  • Thread starter Thread starter PhysicsMark
  • Start date Start date
  • Tags Tags
    Circuit Conceptual
AI Thread Summary
Electrons do not travel down an open branch of a parallel circuit because there is no complete path for current flow, leading to a charge buildup. This accumulation creates an opposing electric field that prevents further electron movement. Electrons cannot escape from a conductor's free end unless subjected to very high voltage, which can cause a spark. The movement of conduction electrons is restricted by Coulomb forces from positive ions, confining them within a "potential well." A complete circuit is necessary for current to flow effectively.
PhysicsMark
Messages
90
Reaction score
0
Hi. I'm not sure if this is the right place to pose this question as it is not homework question.
I'm studying RC circuits in second semester physics and I have a question about the electrons traveling through a circuit. Is it true that electrons won't travel down a branch of a circuit in parallel if it is not connected to anything on one end? If that is true, why do they not travel down those paths?
 
Physics news on Phys.org
Yeah it's true. If the electrons have nowhere to go, then there will be a pile up of charge. Obviously a pile up of charge can't last, since electrons repel. Another way to look at it is that the concentration of charge will result in an electric field that opposes the one that was driving the electrons in the first place. Equalization will happen quickly. You need a complete circuit to have a current.
 
Thank you for the reply.

Does that mean that electrons will continuously accumulate in the broken branch and then scatter?
 
The electrons can not jump out of a conductor at a free end unless it is at very high voltage to make a spark.
The electrons can move freely inside the bulk of the conductor, but kept there by the Coulomb forces of the positive ions. They say that the conduction electrons are confined in a "potential well" inside the metal, and they need a certain amount of energy (Work function, a few eV) ) to escape.

ehild
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging 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