Answer
(a) from one end of its body to the other.
(b) $5.3\times 10^{-9}C$
Work Step by Step
(a) We know that
$V=\frac{Qd}{\epsilon_{\circ}AK}$
This equation shows us that voltage will be higher when d is large and A is small. Thus, it is an end-to-end separation of charge on the body of the fish (rather than side-to-side).
(b) We know that
$Q=\frac{VK\epsilon_{\circ}A}{d}$
We plug in the known values to obtain:
$Q=\frac{(350)(8.85\times 10^{-12})(1.8\times 10^{-2})(95)}{1.0}$
$Q=5.3\times 10^{-9}C$
