Electric eels

$\quad \quad \quad V\quad =\quad E\quad d\\ \Rightarrow \quad E\quad =\quad \frac { V }{ d } \\ \Rightarrow \quad E\quad =\quad \frac { 600 }{ 0.1 } \\ \Rightarrow \quad E\quad =\quad 6000\\$

$\Large{E=\frac{V}{d} = \frac {600~\textrm{V}}{(10~\textrm{cm}/100)~\textrm{m}}=\frac{600~\textrm{V}}{0.1~\textrm{m}}=\fbox{6000}~\textrm{V/m}}$

voltage of the eel is : $600 V$

wide of the eel is : $10 cm$ or $0.1 m$

$E = \frac {F(Electric Force)}{q(charge)}$

$E = \frac {600 V}{0.1 m}$ = $\boxed {6000 V/m}$

In order to solve the problem we have to apply the theorem of gauss.

The flow $\sigma$ is equal to $\sigma=\oint E \dot \delta S = 0$ and to $\sigma=\frac{\sum_{i=1}^{n}Q}{\epsilon_0}$ what we end up is a simple equation of E which is $E=\frac{V}{r}$ due to the fact that $\epsilon_0=\frac{1}{4\pi k}$ and that $V=k\frac{Q}{r}$

There are 6000 Electroplaques, it means every 10 electroplaque produces 1 volt. Total length of eel is 10 cm, that means at every 1 cm there is an electroplaque, which can be treated as a parallel plate capacitor. Now, E = V/d = 600/(10 cm) = 600/(10x10^-2) = 6000

Given that the eel can generate at least 600 V in its body, I assumed that;

H=V/m

Where H is the variable for Electromagnetic Field. If what we are looking for is the strength of its electromagnetic field produce by its body we consider that the assumed equation is to followed.

Substituting the values with the given variables we get;

H= 600 V/ 10 cm

Since, the length is not in its standard form we need to convert it from centimeters to meter

10 cm x 1 m/ 100 cm = 0.10 m, thus

H= 600 V/ 0.10 m

H= 6000 V/m