# Tag Info

6

First, there is nothing wrong with our charge polarity conventions. They are predict electrical phenomena just as accurately as the opposite convention would have done. Did we have a problem if since the begin of their discovery we called them positive particles and negative to protons? We could predict the behavior of electrical phenomena equally ...

4

Electrons in a metal are delocalized, which means they don't know to what atom they belong. The cores are bound in a crystal lattice and the electrons flow around it like a gas, similar to this. When you attach a voltage to both ends of such a material, the electrons bounce into each other and push each other a tiny bit down the wire. Electrons are flowing ...

3

The dissipation in a resistor or wire of the same length $\ell$ doesn't have very much to do with the length - it has to do with the resistance. For a given current (number of electrons per second crossing a particular point), there will be a voltage drop associated with a given resistance - this is Ohm's law. A nice intuitive way to think about this is ...

2

There is a mistake in your working. The surface integral of the Poynting vector $$\int \vec{\Pi} \cdot d\vec{S} = \Pi\ 2\pi a h = \frac{a I^2}{2\pi^2 \gamma a^4} 2\pi a h = \frac{I^2 h}{\pi \gamma a^2}$$ As pwf correctly points out, this power matches the power dissipated as Ohmic heating, $VI$, where $$V I = E h I = \frac{I^2 h}{\pi \gamma a^2}$$ This ...

2

I don't really see what the problem is. If $B$ is zero on the boundary, your equation shows $0= \mu_0 I(t) + \mu_0 \iint \epsilon_0 \frac{\partial \vec{E}(r,t)}{\partial t} \cdot d\vec{A}$, so $$\iint \epsilon_0 \frac{\partial \vec{E}(r,t)}{\partial t} \cdot d\vec{A}=-I(t)$$ Why is it that you expect that this equation isn't satisfied? I can tell very ...

2

A battery is no capacitor, and the actual charge stored in the battery terminals is very low. When you connect the anode of one battery to the cathode of another, that charge is transferred very quickly, and the voltage drops to zero. When you connect anode and cathode of the same battery, a chemical reaction takes place, and charges flow inside the battery ...

2

For your circuit, $V = I\cdot R$. You are plotting (unusually) R along the X axis and $\frac{1}{I}$ along the Y axis, so the slope is $\frac{1}{V}$. Now the fact that this slope is a straight line tells you that the voltage is constant. This means that (over the range of your experiment) your voltage source has a low internal resistance. Imagine for a ...

1

This is a great question! I'm not sure where the power is coming from, but I do know that it corresponds to the power flowing out due to Ohmic heating: Note that $R = \frac{h}{\gamma \pi a^2}$ and $V = IR$, so $E_z = \frac{V}{h}$ (not surprising). Then $\Pi_r|_a = \frac{V I}{2\pi a h}$, and $\oint\Pi\cdot dS = \Pi_r|_a 2\pi a h = VI$, the ohmic dissipated ...

1

If you have current flowing one way through a resistor, then the electrons flow through the other way. Since current flows from the high voltage end of a resistor to the low voltage end, then the electrons come in at the low voltage end and come out at the high voltage end. When electrons (which are negatively charged) go from low voltage to high voltage, ...

1

If it's a simple circuit where Ohm's law applies, then we should get $$V=IR$$ so we see that $$V/I = R$$ $$1/I=R/V$$ $$1/I = (1/V) \times R$$ The gradient should then be $1/V$. Seems like a slightly bizarre plot but if you got a straight line then that makes the maths simple at least!

1

That is a limiting procedure pretty much similar to the definition of derivative. Fix the point $x$ where you want to compute the current density and consider a small disk $D$ centred at that point. There will be a certain current $I(D)$ flowing across this disk. As this disk shrinks on to the point, the ratio $$\frac{\text dI}{\text d ... 1 Take for example a conducting body with an arbitrary shape and volume V connected to a battery that will cause a current flow in the body. You don't know the microscopic structure of the body; so in principle for any point  \mathbf{r} \in V  you can have a different value of I:$$ I=I(\mathbf{r})=I(x,y,z)  Now, take any surface $\Sigma$ in $V$; for ...

1

What I don't understand is the fact that more energy is dissipated within a resistor of length l than would be dissipated in a wire of length l. It just doesn't make sense. You mean, why is it not like this: Electron goes through a resistor while there's a 1 V voltage between the resistor ends -> resistor is heated by energy of 1 V * 1 elementary ...

1

No electrons are being lost; for a current to flow there needs to be a circuit, ie some kind of loop. Also, when current flows, it is not necessarily the electrons themselves that are moving as fast as the current, but rather the signal to move. The copper atom also consists of protons and neutrons in its nucleus, as well as most of its electrons which stay ...

1

Only a few electrons in the outer shell of a copper atom are ever displaced (removed) due to charge forces. The rest all stay in place. So even if copper atoms are mediating a current, the integrity of the atoms themselves is never in doubt.

Only top voted, non community-wiki answers of a minimum length are eligible