Can the electric force lift heavy bodies? We all know the experience that consists of rubbing a plastic pen in one's hair and then using the pen to "animate" some little pieces of paper.
By rubbing the pen, you pull the electrons off it, so it gets a positive charge. And when you approach the pen, which is now charged, close to the pieces of papers, the papers stick together.
I had the same experience with a mosquito. I was able to lift him with the electric force. I used a plastic pen whose the mass is $5.0\times 10^{-3} \;\text{kg}$ to lift a mosquito whose the average mass is around $1.5\times 10^{-6} \;\text{kg}$.
So I calculated (with proportionality) that with a pen weighing around $233 \;\text{tonnes}$, we would lift a man of $70 \;\text{kg}$.
It is possible?
 A: 
So I calculated (with proportionality) that with a Bic Pen weighing
arround 233 tones, we would lift a men of 70kg.

The magnitude of the electrostatic force the pen exerts on the paper, mosquito, or a man, created by rubbing the pen has nothing to do with the mass of the pen.  The force only depends on the net charge acquired at the surface of the pen due to the rubbing. Adding more mass without increasing the rubbing surface area would not increase the net charge and therefore would not increase the electrostatic force.
Whether or not that electrostatic force is capable of causing the paper, mosquito or man to accelerate towards the pen depends on Newtons second law;
$$F_{E}=ma$$
Where $F_E$ is the electrostatic force due to the net charge on the pen  and $m$ is the mass of the paper, mosquito, or man. As others have pointed out, in order to lift a man the strength of the electric field would have to be extremely high.
Hope this helps.
A: It could be possible, but unlikely due to the strength of the electric field involved$^1$.
The electric field is proportional to the amount charge and does not depend on the mass of any of the bodies.
If you wanted to lift a man who's mass is $70kg$ using electrostatic forces,  you would need the electrostatic force to be equal to his weight, or $$F_E=70g\approx 700N$$
$^1$ The electric potential difference generated by such a huge collection of charge, may exceed the breakdown voltage for air. At that point the charge will rapidly cross the boundary between the two objects and quickly neutralise the system.
A: In principle it is possible but in practice it is extremely difficult.
The problem is that the field strength required to lift any significant weight is extremely high and arcing or breakdown takes place, collapsing the field.
The principle was once studied for hovercraft and VTOL aircraft, but the voltage needed to attain any height was extreme, the field gradient before breakdown still allowed only modest weights, and even if it could have been achieved the risk of electrocution to anybody underneath would have been far too great. I recall an SF story with such vehicles in, but the idea stopped there.
What has been achieved is the use of electrostatic forces to accelerate air, creating thrust for ion-propelled aircraft. Light and flimsy VTOL models with vertical ion thrust have lifted-off and hovered, but they trailed wires back down to the heavy voltage generator on the ground. MIT also recently made a model fixed-wing plane, the MIT EAD Airframe Version 2, sustain flight, with onboard horizontal ion thrust generation and more efficient conventional aerodynamic lift. But the model did not have surplus thrust for takeoff, it had to be launched.
A: A relatively massive object has been levitated with electrostatic forces.
Gravity Probe B used fused quartz spheres about the size of ping-pong balls. These were designed not only to be centred by electrostatic forces in orbit, needing only fractions of a volt, but also to be levitated in the lab in 1g gravity for testing. This needed about 1500 V across the 0.001" clearance between sphere and housing electrodes, in a vacuum of course.
I'll see if I can find some links to some design papers, please feel free to beat me to it!
A: Unless I miscalculated, this does not seem possible for the following reason.
Let us assess the electric field required to suspend a man with a mass of $m=70kg$. For simplicity, let us consider a plane capacitor with the plate area of $A=1m^2$, which seems reasonable for "one side of a human body". The force $F$ between the plates of the capacitor is $\frac{\epsilon_0 A V^2}{2 x^2}=\frac{\epsilon_0 A E^2}{2}$, where $\epsilon_0=8.85\cdot 10^{-12}F/m$, $V$ is the potential difference between the plates, $x$ is the distance between the plates, $E$ is the electric field inside the capacitor. If $F=mg\approx 686N$, we obtain $E=\sqrt{\frac{2 m g}{A\epsilon_0}}\approx 1.24\cdot 10^7 V/m$, whereas electric breakdown in atmospheric air takes place when electric field exceeds $3\cdot 10^6 V/m$.
EDIT(7/30/2021): @EdL asked good questions in a comment:

So what would happen instead, would the capacitor discharge through
the air?

Yes.

Could you do the same electrostatics question with the mosquito to
compare what field is required for that?

So let us assume that the mosquito has the following size and weight: 4mm and 5mg. Let us assume that the mosquito has a square shape in a plan (this may look silly, but one can make a different assumption and calculate - note the square root in the formula), so the area is $1.6\cdot 10^{-5}m^2$. Substituting this data in the above formula, one obtains $E\approx 8.32\cdot 10^5 V/m$.
