Folding a charged triangle.

Dice the original triangle into many tiny pieces that are essentially point charges with charge $q$ . The electric potential energy of the original triangle would be the sum of the electric potential energy of all pairs of little pieces: $\frac{1}{2}\displaystyle \sum_{j=1}\displaystyle \sum_{i=1}kq^2/r_{ij}$ where $k$ is Coulomb's constant and $r_{ij}$ the distance between pieces $i$ and $j$ . Let's call the value of this sum $X$

The first folding is equivalent to reducing the distance between each pair of charges by a factor of $\sqrt{2}$ , so the energy after the first fold is $\sqrt{2}X$ .

$\sqrt{2}X-X = 1 J$

$X = 3.4142... J$

The second folding brings every pair closer again by a factor of $1/\sqrt{2}$ so energy difference before and after the second folding is given by:

$2X- \sqrt{2}X = W_2$

$2(3.4142)- (\sqrt{2})3.4142 = W_2$

$2J = W_2$

We can compute the work required to fold a right isosceles triangle by integration. For a triangle of height h and charge Q the answer should be of the form $W= C \frac{Q^{2}}{h}$ where C is an unknown constant. The problem becomes trivial when one realizes that C has the same value for all right isosceles triangles folded as in this problem. Note that the total charge of the triangle does not change when you fold it, i.e.
$Q_{1}=Q_{2}$ and therefore we can write $\frac{W_{2}}{W_{1}}= \frac{h_{1}}{h_{2}}.$
Basic trigonometry yields $\frac{h_{1}}{h_{2}}=\sqrt{2}$ and we arrive at the result $W_{2}=\sqrt{2} W_{1}.$

I assume that we have here a process which requires energy dissipated in the folding line of the dielectric material. And has nothing to do with it's dielectric properties:)

The 1Watt folding energy was dissipated on a folding line of a length(red broken line) computed as follows: Sqrt(a^2-b^2) where b=(sqrt(2)xa)/2= half length of the Hypotenuse of the leftmost triangle. The folding length thus is = a/sqrt(2) Using the Pythagorian theorem on the folded triangle(middle triangle) where the folding side is along the height(broken red line) . The Hypotenuse is (Sqrt(2)xa)/2 the base is a/2 which gives a height of a/2, this is the length of the folding line on which bending energy is spent. But this time we have to fold a two layered triangle, because of the previous folding, so the effective bending length will be 2xa/2=a, which if divided by a/Sqrt(2) length for 1watt spending will give us Sqrt(2)~ 1.4142watts about

work done in shearing along a length of say l is proportional to area lt where t is thickness of plate.In second fold,shear length is square root 2 times the first case.