Posts from Quiz Auto-translated

There is no answer to my question yet.

It won't make it. It will get stuck in the dielectric between the plates.

An electron with a speed of 100 m/s enters a capacitor through a hole in the negative plate.
The electric field strength in the capacitor = 3.8*10^(-6) N/C
The distance between the plates is 1 mm
The electron has a mass of approximately 9*10^(-31) kg and a charge of 1.6*10^(-19) C.
How deep into the capacitor can the electron fly?

The electron will experience a force equal to the electric field strength multiplied by its charge. This force will slow it down. To find the deceleration, this force must be divided by its mass. Furthermore, the distance it travels before stopping will be equal to the square of its initial velocity divided by twice the deceleration. However, it certainly cannot travel further than the distance between the plates. I will now calculate the values and try to create a diagram.

Added 2 minutes ago
The field strength is directed from the negative plate to the positive plate, and since the electron's charge is negative, the field will slow it down.

Added 25 minutes ago
The force that will slow down the electron: F = 3.8*10^-6*1.6*10^-19 = 6.08*10^-25 N
The deceleration will be: a = F/m = 6.08*10^-25/9*10^-31 = (6.08/9)*10⁶ m/s²
And the distance will be: s = v²/(2a) = 100²/(2*(6.08/9)*10⁶) = 10⁴/(10⁶*2*6.08/9) = 9/(100*2*6.08) = 9/1216 = 0.0074013157895 m. In short, it will reach the opposite plate :)

I made the same statement, but without the calculations. And it was incorrect.

It will get stuck in the dielectric between the plates.

Since nothing is said about the dielectric, its presence is not implied.

Can an electron fly in?

This question was already asked. Obviously, the presence of a dielectric at least does not help this "can," so there is no need to consider the dielectric.
Unfortunately, there is still no answer to the question.

Actually, I was wrong. All my arguments would be relevant in the case where an electron would fly through the positive plate... because, in reality, the electric field is directed from the positive plate to the negative plate, so the field would also accelerate it in this capacitor, and it would reach the opposite plate, and its speed would be greater. But even if it were to fly through the positive plate at that speed, it would still reach the negative plate, as in such a field, it could travel more than 7 millimeters in its initial direction, and the distance between the plates is only 1 mm.

Added 2 minutes later
Well, 1 millimeter is the answer.

It seems like it won't even fit inside the capacitor...

It seems like it won't even enter the capacitor...

No, it will fly through it from one plate to the other, and it doesn't matter which plate it enters through. Even if it entered with such speed through the positive plate, it would still reach the negative plate.

Added after 42 seconds
Well, half a millimeter, if it goes deep, if it goes to the very center of the capacitor.

Who knows. I don't remember much from 9th-grade physics, and I don't feel like looking for a reference book.

Well, 1 millimeter – that’s the answer.

Still no.

Well, half a millimeter, if going deeper, and if the deepest point is the middle of the capacitor.

Yes, the answer is correct. It's just that, terminologically, you can't go deeper into a one-millimeter object than half a millimeter. After going half a millimeter deep, the electron will start moving towards the opposite surface.)
A more obvious riddle: "How deep into the forest can a deer run?"
Hermit, please, ask the next riddle.

The force that will slow down the electron: F = 3.8 * 10-6 * 1.6 * 10-19 = 6.08 * 10-25 N
The deceleration will be: a = F/m = 6.08 * 10-25 / 9 * 10-31 = (6.08/9) * 106 m/s2
And then we'll calculate the distance: s = v2 / (2a) = 1002 / (2 * (6.08/9) * 106) = 104 / (106 * 2 * 6.08/9) = 9 / (100 * 2 * 6.08) = 9 / 1216 = 0.0074013157895 m

By the way, I chose the numbers so that the result would be 0.75 mm if you start solving such a problem. I must have made a mistake by an order of magnitude somewhere, I was calculating "in my head." Never mind, it doesn't matter for the solution; the numbers are symbolic.

Riddle

Due to the abnormally high temperatures expected in the city within 24 hours, 240 tanks of kvass were brought in. Five criminals poisoned one of the kvass tanks; they were all caught and sentenced to death... They don't remember which tank they poisoned... Is it possible, by experimenting on the convicted, to identify the poisoned kvass tank, if it is known that even a negligible dose of poison is fatal, and the poison begins to take effect within 12 hours of entering the body?

The answer must be justified. An answer without justification will not be accepted.

Amendment: the poison will definitely kill the convicted person within 12 hours, but it may start to take effect at different times.

It depends on how accurate the 12-hour figure is. You can water them every minute, but the specifics of their physiology and the accuracy of the instruments won't allow you to establish the truth.
Update: So, we divide 240 by 5 and by 12. This means they need to be watered every 4 minutes, marking the tank.

The heat will be terrible. If we don't start giving the townspeople kvass to drink within 24 hours, it will be a disaster... If the perpetrator drinks even a drop of kvass from the poisoned tank, the poison will kill him within 12 hours!

So, we'll analyze the first 12 hours and then see the results. We can increase the time between analyses to improve accuracy, but then we'll have to sacrifice the number of tanks we can check.