The pressure will be the same.
Just look at the ideal gas law:
PV = nRT, here P is pressure, V is volume, n is the number of molecules and T is the temperature. R is just a constant.
Solving for P you get:
P = nRT / V
Now, right after you close the lid on each container, notice that the...
DaleSpam:
Strictly speaking, you are correct of course. However, if the magnitude of the electric field X at every point between point A and B were larger than the magnitude of the electric field Y on every point between point C and D, it's common to refer to field X as the "stronger electric...
Air resistance, which is a kind of friction. When an object falls through the Earth's atmosphere, it collides with molecules in the air. It slows the feather down more than the bowling ball, because the feather has a larger surface area in comparison to its mass.
Well, that's all I've been trying to point out for the latter half of this thread. A voltage "by itself" is not enough to determine the strength of an electric field between two points, you have to also know the distance between the two points in question.
But anyway, I think I understand this...
"For a conservative field a larger potential drop along a path does, in fact, automatically imply a larger field along the path."
Yes, if it is the same path. But if you have the same potential difference between the endpoints of a wire with length L, and a wire with length 2L, shouldn't the...
Adding the third point was only to illustrate that a larger voltage does not automatically imply a stronger electric field.
Another example: If all you know is that person X drove 100 miles, and that person Y drove 50 miles, it does not say anything about their velocities, even though velocity...
Well you're right, there can't be any prolonged current in such a wire, but I just imagined it in the first nanosecond as the charges started to rearrange themselves to oppose the outside field.
For some reason I always find the most basic things to be the hardest to understand. I can do...
I think the distinction still applies. I imagine a conductor of some length l, placed in a uniform electric field.
If you take three points a, b and c on the conductor, where v(a) < V(b) < v(c), then V(c) - V(a) > V(b) - V(a).
If the current is larger at point c, then it really is the same...
Voltage = potential difference.
As for the water analogy, you could say that pressure difference drives the water, but it is the gravitational field thas is responsible for the pressure in the first place.
I don't think the analogy is perfect either, as the significant gravitational field is...
I'll admit to still being a little confused about this. The whole point of the thread was that I couldn't understand "why" voltage is the supposed driving force of current, as it is the electric field which asserts a force on a charge (and thereby causes it to move).
As you pointed out, the...
jartsa:
I didn't notice your reply, so I wasn't replying to you! What you wrote made sense though. The initial problem was that I saw voltage described as the driving force for current in several places. I then pictured some cases where higher voltage meant lower electric field, or the same...
You're right, I meant for a negatively charged plate. It doesn't really matter where I define the voltage to be zero, the difference in voltage between two points will come out the same.
Anyway, the point of my question was to understand why higher voltage equals higher current. I get that you...
Now I'm spamming my own tread, but say you have a charged plate with infinite area. The force is the same everywhere, while the potential is E times the perpendicular distance away from the plane. So even if you increase the voltage by moving away, it doesn't increase the strength of the...
Won't that be because the force of friction works in the opposite direction of the total movement, not seperately for the translational and rotational?
What I mean is, if you take one point on the corner of the spinning book, it has a movement due to the rotation around the center of mass, and...
Perhaps I'm just really stupid... but I thought some more about this, and now it does not make any sense again. The electric field is the "negative" of the gradient of the voltage, so the voltage does increase with distance (though not in magnitude).