From where the E inside the battery comes from?

  • Thread starter Thread starter kashan123999
  • Start date Start date
  • Tags Tags
    Battery
AI Thread Summary
The electric field inside a battery is established due to the redox chemical reactions occurring within it, which move ions to their respective terminals. This process creates a potential difference and an electric field before any external circuit is connected. Work is required to move positive charges from the negative to the positive terminal against this electric field, building up voltage. Once charges accumulate, current flows from the positive to the negative terminal, following conventional current flow. Understanding these principles helps clarify the origin of the electric field in batteries.
kashan123999
Messages
98
Reaction score
0
So as I am junior high school student just learned that when the terminals of wire are connected to a battery,an E (electric field) is established inside the battery from positive to negative end/terminal (How did I assumed one as positive and one as negative? COnvention regarding electric fields,am i right?),Work is required to push say,positive charges from negative terminal to positive terminal i.e against the electric field (WHY? obviously to build potential/potential difference b/w both ends) so the work is done by the energy provided by the battery,When positive end would be have charges and negative end do not have them...obviously the current would start to flow from positive to negative terminal...all of this is actually conventional current..

My question is,from where the Electric field inside the battery comes from?
 
Physics news on Phys.org
kashan123999 said:
So as I am junior high school student just learned that when the terminals of wire are connected to a battery,an E (electric field) is established inside the battery from positive to negative end/terminal (How did I assumed one as positive and one as negative? COnvention regarding electric fields,am i right?),Work is required to push say,positive charges from negative terminal to positive terminal i.e against the electric field (WHY? obviously to build potential/potential difference b/w both ends) so the work is done by the energy provided by the battery,When positive end would be have charges and negative end do not have them...obviously the current would start to flow from positive to negative terminal...all of this is actually conventional current..

My question is,from where the Electric field inside the battery comes from?

The EMF is present inside the battery before any external circuit is connected. Have a read through this introductory article at wikipedia:

http://en.wikipedia.org/wiki/Battery_(electricity )

:smile:
 
Last edited by a moderator:
  • Like
Likes 1 person
berkeman said:
The EMF is present inside the battery before any external circuit is connected. Have a read through this introductory article at wikipedia:

http://en.wikipedia.org/wiki/Battery_(electricity )

:smile:

thank you i would surely read it as well :)
 
Last edited by a moderator:
The redox chemical reaction moves the ions to their respective terminals creating an electric field and potential difference. Once charges accumulate at the terminals, the redox process is propelling ions to move against the electric field. This increases the voltage at the terminals (potential difference) as well as increase the E field energy. Did I help?

Claude
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Thread 'A scenario of non-uniform circular motion'
(All the needed diagrams are posted below) My friend came up with the following scenario. Imagine a fixed point and a perfectly rigid rod of a certain length extending radially outwards from this fixed point(it is attached to the fixed point). To the free end of the fixed rod, an object is present and it is capable of changing it's speed(by thruster say or any convenient method. And ignore any resistance). It starts with a certain speed but say it's speed continuously increases as it goes...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top