Understanding Fundamentals of Electronics: Solving an Electric Charge Problem

[vasya]

Homework Statement

How many electrons will flow into the cube, connected to the 1.5v battery negative terminal, with positive being grounded

Relevant Equations

i dont know

It's not a homework. I came up with this problem myself. Trying to understand fundamentals of electronics. Do you know how to solve it? Is voltage somehow related to electron energy levels? What knowledge should I gain to be able to solve problems like that? Thank you!

If we ground the cathode (+) of a battery with, for example, 1.5v, and connect the anode (-) to an iron cube of 1 cm³

(26 protons in the nucleus,
density - 7.874 g/cm³
Atomic mass - 55.845 (g / mole)
chemical count - 0.1409 mole
atoms - 0.848 *10^22
protons - 22.06 *10^22 and the same number of electrons with zero charge)

How many electrons will flow into the cube, or what will be the charge in it?

ps: the electrostatic field of the Earth does not change, so it's not a capacitor, right?

By the way, if we will apply very high voltage to the cube, electrons will start to jump out of crystal lattice and fly away right?

 

[BvU]

Hello @vasya,

$\qquad$!​

Good questions, not so easy to answer at an appropriate level (which level that is in your case, isn't really known to us).

There is a relationship between charge and voltage$$Q = CV$$ with Q the charge, V the voltage and C the so-called capacitance. For the shape of the cube it's rather difficult, but for a sphere the capacitance can be shown to be $C = 4\pi\varepsilon_0 R$, so that we get $$Q=4\pi\varepsilon_0 R \, V\ .$$A 1 cm³ sphere has a radius of 6.2 mm, so with 1 V we get a charge of 6.9 10¹³ Coulomb, which is about 4300 electrons.

[Edit]yes, 6.9 times ten to the minus 13 Coulomb, of course. Sorry.
at first I had the 'Coulomb' inside the exponent and concentrated on fixing that...$\$

 

[hutchphd]

Homework Statement:: How many electrons will flow into the cube, connected to the 1.5v battery negative terminal, with positive being grounded
Relevant Equations:: i don't know

How many electrons will flow into the cube, or what will be the charge in it?

ps: the electrostatic field of the Earth does not change, so it's not a capacitor, right?

By the way, if we will apply very high voltage to the cube, electrons will start to jump out of crystal lattice and fly away right?

This will have capacitance. Roughly it will be given by $C= {4\pi{\epsilon}_0}\frac a 2$ where a is side of cube. Then Q=VC

Given enough voltage (the work function) the electrons will be ejected from the metal. Usually they will bounce back but If this happens at a high enough rate and energy to cause ionization of the air a spark cascade will develop.

 

[vasya]

thank you all!

 

[Delta2]

Interesting problem you came up with, considering that you now beginning your studies to electricity/electromagnetism.
In my opinion both of the above answers are very good, just to ask @hutchphd , the formula for the cube must have a typo, it must be $C=4\pi\epsilon_0\frac{a}{2}$ and to emphasize that it is an approximation and not exact as in the case of a sphere.

Also to add that for your problem we must consider some assumptions:

1. That the grounded electrode has exactly 0V potential
2. The battery is ideal, and can maintain always 1.5V potential difference between the electrodes
3. The electrodes behave as ideal conductors.

 

[Delta2]

A 1 cm3 sphere has a radius of 6.2 mm, so with 1 V we get a charge of 6.9 1013 Coulomb, which is about 4300 electrons.

It must be $6.9\times 10^{-13}$ right? (negative exponent I mean)

 

[hutchphd]

That the grounded electrode has exactly 0V potential

I don't understand this requirement .
Thanks for the correction to my original post...I saw how badly I screwed it up immediately and think I corrected it before yours dropped. (Thought I had gotten away un-noticed.l...!)

 

[Delta2]

I don't understand this requirement .

Yes well i would better say that "The potential of each electrode is not affected by the charge of the other electrode".

 

[bob012345]

Here is a paper on the numerical capacitance of a unit cube.

https://www.researchgate.net/publication/239582798_UNIT_CUBE_CAPACITANCE_CALCULATION_BY_MEANS_OF_FINITE_ELEMENT_ANALYSIS

 

[nasu]

.I saw how badly I screwed it up immediately and think I corrected it before yours dropped. (Thought I had gotten away un-noticed.l...!)

The number of electrons too needs a correction.

 

[bob012345]

Also, it would be interesting to do a capacitance measurement of a 1 cm cube vs. a similar sphere at some DC voltage. Since it would not by a typical 2 pole capacitor, any suggestions as how to do it? My initial guess would be to hook the negative terminal of a battery to the electrically isolated cube or sphere and the other terminal to ground. Would the capacitance be measured en between the cube or sphere and ground? Fluke digital multimeters have a capacitance setting and I was thinking of using that.

 

[bob012345]

Correcting an earlier error I made, the average distance from the center to an edge of a unit cube is actually 0.640395!

This suggests $C≈4 \pi \epsilon_0 (0.6404) a$ where $a$ is the cube length.

The paper (no full access but the results are in the abstract) referenced below computes capacitance for a cube as; $$C=4 \pi \epsilon_0 (0.6606±0.0001) a$$
https://www.sciencedirect.com/science/article/abs/pii/S0304388604001044#:~:text=Our final result for the,therefore the best available value.

Another paper (check out the neat 3D plot of surface charge distribution!) with similar results;

https://www.cs.fsu.edu/~mascagni/papers/RIJP2003_6.pdf