capacitanceelectromagnetismelectrostaticspotentialvoltage
Suppose to have two capacitors in series:
The voltage in the middle point will be:
$$
V_X = V_1 \frac{C_1}{C_1+C_2}
$$
How can this be explained? It's been asked in electronics, and explained in terms of impedence and charge equality, but none of the explanations is satisfying to me, as I think it should involve charge conservation (Gauss theorem?) and/or electric fields/potentials.
Could you enlighten me?
then why is there no potential difference between the two capacitors
It's not quite clear what you mean here but do understand that charged capacitors are electrically neutral.
When a capacitor is "charged", it is not electrically charged, it is energy charged in the same sense as when we say a battery is charged.
There is nothing mysterious about two series connected circuit elements having different voltage drops. Think of two series connected resistors with different resistor values.
I would have thought that as plate B is positively charged and plate C is negatively charged, there would be a potential difference between the two
You're forgetting something fundamental: The plates B and C along with the wire that connects them are conductors. But, for an ideal conductor, charge distributes itself so that there is no (static) potential difference across the conductor.
The voltage between the bottom plate of C1 and the top plate of C2 is zero precisely because a conductor connects the two plates.
Here's what I do (or rather, what I just did) to convince myself of your result.
We're trying to "reduce" the two capacitors in series to an equivalent capacitor. Equivalent in every way, including how much charge would flow when discharging the capacitor.
Imagine discharging the two capacitors that are series. How much total charge would flow? Since, as you've stated, the "input" and "output" sides store equal magnitude charges $|Q_1|=|Q_2|\equiv Q$, then we should be able to recover this much charge during discharging.
By discharging the equivalent capacitor, then, one should expect to have a total charge $Q$ flow during discharging. Thus, $Q_\mathrm{equiv}=Q$.
Best Answer
Suppose you imagine the battery to be a variable voltage, and start with the voltage at zero. Obviously everything is uncharged.
Now turn the battery up to 1V. As you do this positive charge leaves the positive terminal and an equal and opposite negative charge leaves the negative terminal. We know the charges leaving the positive and negative terminals must be the same because the battery is a conductor and can't develop a net charge like a capacitor. Let's call the charge that leaves the battery $Q$.
The only place the charge that leaves the battery can go is onto the capacitors, so both capacitors now have a charge of $Q$ on them. We know that for a capacitor of capacitance $C$, the voltage across the capacitor is given by:
$$ V = \frac{Q}{C} $$
Call the voltage of the top (1$\mu$F) capacitor $V_1$, and the voltage of the bottom (2$\mu$F) capacitor $V_2$. Then:
$$ V_1 = \frac{Q}{C_1} $$ $$ V_2 = \frac{Q}{C_2} $$
Dividing the first equation by the second plus a bit of quick rearrangement gives:
$$ V_1 = \frac{C_2}{C_1} V_2 $$
The two voltages must add up to 1V because we have a 1V battery, therefore:
$$ V_1 + V_2 = 1$$
If you substitute for $V_1$ you get:
$$ \frac{C_2}{C_1} V_2 + V_2 = 1$$
and dividing through by $(1 + C_2/C_1)$ gives:
$$ V_2 = \frac{1}{1 + C_2/C_1} $$
Tidy this up by multiplying to top and bottom of the right hand side by $C_1$ and you get the equation you're trying to prove:
$$ V_2 = \frac{C_1}{C_1 + C_2} $$
Just to check, feed in $C_1 = 1$ and $C_2 = 2$ and $V_2$ does indeed come out as 1/3V.