Question

The 9-inch-long elephant nose fish in the Congo River generates a weak electric field around its body using an organ in its tail. When small prey (or even potential mates) swim within a few feet of the fish, they perturb the electric field. The change in the field is picked up by electric sensor cells in the skin of the elephant nose. These remarkable fish can detect changes in the electric field as small as 3.0 μN/C. (a) How much charge (modeled as a point charge) in the fish would be needed to produce such a change in the electric field at a distance of 75 cm? (b) How many electrons would be required to create the charge?

Answer 1

Answer:

(a) 1.88*10^-16 C. (b) 1172 electrons

Explanation:

An electric field (E) can be defined as a space surrounding a charged object (Q) that exerts force on all charged particles around the object. Mathematically:

$E = \frac{kQ}{d^{2} }$  

Where: E is the electric field = 3*10^-6 N/C, d is distance = 0.75 m and k is the proportionality constant = 8.99*10^9 N*m^2/C^2.

Thus:

(a) Q = E*d²/k

Q = (3*10^-6)*(0.75²)/8.99*10^9 = 1.88*10^-16 C

(b) An electron has a charge of 1.602*10^-19 C. Therefore, the number of electrons that would need 1.88*10^-16 C is (1.88*10^-16)/(1.602*10^-19) = 1171.7 . That is approximately 1172 electrons.