Question

Guiana dolphins are one of the few mammals able to detect electric fields. In a test of sensitivity, a dolphin was exposed to the electric field of charged electrodes. The electric field was measured by detecting the potential difference between two electrodes located 1.0 cm apart along the field lines. The dolphin could reliably detect a field that produced a potential difference of 0.50 mV. What is the corresponding electric field strength?

Answer 1

Answer:

0.05 V/m

Explanation:

For a uniform electric field, electric field strength and potential difference are related by

$E=\frac{\Delta V}{d}$

where

E is the electric field strength

$\Delta V$ is the potential difference

d is the distance between the two points

Here we have

$\Delta V= 0.50 mV=5\cdot 10^{-4}V$

$d=1.0 cm=0.01 m$

So, the electric field strength is

$E=\frac{5\cdot 10^{-4} V}{0.01 m}=0.05 V/m$