Weight difference between an empty and a full memory stick

[mheslep]

Initially I made a guestimate based on typical structure of modern flash memory of about 10^3 electrons per floating gate (that is, per bit). Stupidly I took the mass increase to be the mass of these electrons (approx 10^(-27) kg per bit). That's nonsense of course, as each cell remains overall charge neutral and the electrons are just redistributed from one plate of the capacitor to the other.

Looking at it again I'll say approx $n\, q_e\, V\, /\,(4\,c^2)$ kg per bit. So based on n approx 10^3 electrons per bit and assuming V is a few volts, I get about 10^(-33) kg per bit as a serious guestimate for modern flash memory.

I speculate you were more on the right tract in the first paragraph. When we say neutral, yes that means that at the macro scale of the memory stick the charge will be 'neutral', but not necessarily to every last −1.602×10−19 Coulomb (e), not over the many moles of electrons in question here. It could be that some state of the stick could well carry a net charge of a few e. Normally the mass of a few electrons would be insignificant, but compared to the minuscule E/c^2 mass grasped at in the second paragraph, the rest mass of even a few electrons is king.

 

[jacksonwalter]

Didn't have time to read all 8 pages, hope this isn't redundant. Usually the "1's" are stored as charged capacitors, whereas the "0's" are stored as uncharged capacitors. A charged capacitor is going to have energy = 1/2*C*V^2, therefore mass = E/c^2.