Navigation of electric fish. Certain fish, such as the Nile fish (Gnathonemus), concentrate charges in their head and tail, thereby producing an electric field in the water around them. (See Figure $19.67 . )$ This field creates a potential difference of a few volts between the head and tail, which in turn
causes current to flow in the conducting seawater. As the fish swims, it passes near objects that have resistivities different from that of seawater, which in turn causes the current to vary. Cells in the skin of the fish are sensitive to this current and can detect changes in it. The changes in the current allow the fish to navigate. (In the next chapter, we shall investigate how the fish might detect this current.) Since the electric field is weak far from the fish, we shall consider only the field running directly from the head to the tail. We can model the seawater through which that field passes as a conducting tube of area 1.0 $\mathrm{cm}^{2}$ and having a potential difference of 3.0 $\mathrm{V}$ across its ends. The length of a Nile fish is about $20 \mathrm{cm},$ and the resistivity of seawater is 0.13$\Omega \cdot \mathrm{m} .$ (a) How large is the current through the tube of seawater? (b) Suppose the fish swims next to an object that is 10 $\mathrm{cm}$ long and 1.0 $\mathrm{cm}^{2}$ in cross-sectional area and has half the resistivity of seawater. This object replaces the seawater for half the length of the tube. What is the current through the tube now? How large is the change in the current that the fish must detect? (Hint: How are this object and the remaining water in the tube connected, in series or in parallel?)