Could a variable capacitor divider replace a Variac? Hmmm... You can definitely drop down the voltage, and ideal capacitors don't dissipate any power. So it seems, at first glance, that you could use a capacitor divider as a lossless voltage step-down device for AC.
So you could use a cheaper variable cap divider as a replacement for a Variac? And it would be continuously variable, too, whereas the Variac is only variable in discrete windings.
There has to be something wrong with this. What am I not thinking of?
 A: I don't think you've considered the output impedance of such a divider.
Variable capacitors typically have relatively small capacitance which means that, at AC mains frequency, the output impedance will be enormous compared to the variac, i.e., your variable capacitor divider will only work with (very) high-impedance loads.
In other words, you will not able to deliver significant power this way.
A: Have you considered some kind of small inductor push pull, pull down/flyup system, you get your 50/60/400Hz by pulse width modulation, you can change to antiphase  by switching to an alternative switch at the other end of your inductor assembly.  Can't say much more or previous employers may come down on me like a ton of bricks.
Operating at a higher frequency you can keep the size of the inductor down.  Though unless FETs have improved in recent years you may need to have some ultra fast rectifiers in series with your FETs to keep dissipation and RFI manageable.
As to those who say it's not physics, let's face it, almost every difficult question in electronics invoves physice.  Charge storage, signal propagation, radiation of energy, come on guys.
A: Let's do the circuit analysis. I believe you are proposing a pair of series capacitors, with their junction being the point where you take a (resistive) load - something like this:

The (AC) output impedance of this circuit is given by the parallel capacitance of $C_1$ and $C_2$, and is
$$Z = \frac{1}{j\ \omega\ (C_1+C_2)}$$
We can calculate the power you can develop - and the voltage range you can reach. If we assume that $C_2$ is variable from 0 to $2C_1$, we can plot the voltage across $R_1$ and the power developed:

In this case, when $\rm C_1 = C_2 = 1 mF$, the power dissipated in $R_1$ when the voltage is 1 V is 2.5 mW; in other words, the circuit works as expected.
Bottom line:


*

*it is hard to get the voltage to both rails when you have just a single variable capacitor

*the control works well as long as $\rm RC\gg \frac{1}{\omega}$, so that the capacitors act like a dividing network without imposing significant lag (their impedance is small compared to the load). This is EXTREMELY challenging to do for any real loads at 50 or 60 Hz, as variable capacitors typically have very small capacitance (certainly not 1000 µF like I was assuming in the above).


For your interest, the following lines of Matlab generated the plot: you can play with the values and see what happens for a "realistic" scenario of load and capacitance.
C1 = 1e-3;
C2 = linspace(1,200,200)*1e-5;
R = 100;
w = 60*2*pi;
Z1 = 1./(1j*w*C1);
Z2 = 1./(1/R +1j*w*C2);
V = Z2(:)./(Z1+Z2(:));
figure; 
subplot(2,1,1)
plot(C2./C1, abs(V))
xlabel 'C_2/C_1'
ylabel 'Voltage ratio'
I = V/R;
P = abs(V.*I);
subplot(2,1,2)
plot(C2./C1, 1000*P)
xlabel 'C_2/C_1'
ylabel 'Power (mW)'

