The electrical energy you pay for doesn't just go into the appliances; some amount of it ends up as heat in the electrical path from the meter to the loads. Also, heat from a load such as a heating element can travel along the electrical conductors away from the application (the water tank) - remember: most good electrical conductors are also good thermal conductors. So, your scaled up water heater element will still produce the same heating power as it always did, but less of the heat ends up in the water, and more ends up outside the water heater. Some of that wasted heat escapes along the wiring, and, because of the increased duty cycle to compensate for the ineffiency, the wiring spends more time carrying current and wasting energy.
Bottom line: energy is conserved in that it doesn't simply disappear... it just gets misplaced to become waste heat somewhere and provides no benefit.
Often the easiest way to do such problems is using potential rather than the electric field. And always, it is best to use the appropriate coordinates. In this case, cylindrical coordinates with the rod at $\rho=0$ and extending from $z=-L/2$ to $z=+L/2$ are natural.
So what you do is say that for an element of rod with a tiny length $d\ell$ and charge $\frac{Q}{L}d\ell$ located at $0,z,\theta$ (where $\theta$ is actually moot since $\rho = 0$), the potential at any other point $(r,\zeta, \theta)$ os
$$\frac{Q}{L\sqrt{r^2 + (z-\zeta)^2}}d\ell$$.
Then the potential due to the whole rod is
$$ V(r,\zeta, \theta) =\int_{z=-L/2}^{+L/2} \frac{Q}{L\sqrt{r^2 + (z-\zeta)^2}}d\ell$$
Then you do the integral, and when you get $V(r,\zeta, \theta)$ you take its gradient, because the negative of the gradient will be the electric field.
CAREFUL: The gradient operator in cylindrical coordinates is not quite just $(\frac{\partial}{\partial \rho},\frac{\partial}{\partial z},\frac{\partial}{\partial \theta})$. Learn what it is, then apply it to your $V$. And don't foregt to convert the field back to get $(E_x, E_y, E_z$ if your problem asks for those.
To learn about cylindrical coordinates, wikipedia is your friend
Best Answer
This device looks like a inherently bad idea safety-wise, for the reason you found. I don't know what exactly is inside the handle, but you have to assume all it is doing is connecting wall power to a resistive heating element.
I imagine the outside of the heating element is intended to be insulated from everything else. (By the way, this has nothing at all do to with its ability to make heat. In fact, the heating element can't be made from a insulator else its resistance would be high and no heat would be generated.) Even if that insulation is intact and working, there could be some capacitive pickup from the input power to what appears to be a metal jacket of the heating element. This is what you may have felt.
However, if water got into the handle part, then all bets are off. It is probably sealed, at least enough to look sealed or pass some sort of test when new. I would be very careful to never let the handle part get wet anyways. Stuff happens. Does the power plug have two prongs or three? If three, maybe the metal jacket around the heating element is grounded. That would help in most cases, but can also make things worse in other cases.
Clearly something is already not right with this device if you felt a shock. I would ditch it, find a proper way to heat water, and move on. If think Murphy's law and the laws of physics don't apply to you, then at least only use this device on a circuit with a ground fault interruptor or with a separate isolation transformer.