I Where does the formula I = -e/T comes from?

AI Thread Summary
The formula I = -e/T relates to electric current, defined as the rate of charge passing a point. In this context, 'I' represents current, 'Q' is charge, and 't' is time, leading to the equation I = Q/t when the rate is constant. The term '-e' denotes the charge of a single electron, indicating that the formula describes the flow of electrons. Thus, the equation quantifies the rate at which an electron, or a charge equivalent to that of one electron, moves past a specific point. Understanding this relationship is crucial for grasping the fundamentals of electric current.
Syazani Zulkhairi
Yeah, where does it comes from?
 
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You will have to be more specific and provide context.
 
I'm going to assume your talking about electric current?

In that case, electric current is defined to be the rate of charge passing a particular point.

Charge is given the symbol Q, time t. So if the rate is constant, the current I = Q/t.

In terms of your equation, -e is the charge of one electron, so your current would represent the rate at which a single electron (or something with that amount of charge) passes a particular point.
 
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Thread 'Gauss' law seems to imply instantaneous electric field'

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...
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Thread 'Do electron density waves accompany EM waves in coaxial cables?'

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...
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