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 3.00 μN/C. 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 99.5 cm?

How many electrons would be required to create that charge?

Answer 1

Part A.

Electric field is given by:

E = k*Q/R^2

Given that

E = 3.00 uN/C = 3*10^-6 N/C

k = electrostatic constant = 9*10^9

R = distance = 99.5 cm = 0.995 m

So,

Q = E*R^2/k

Using known values:

Q = 3*10^-6*0.995^2/(9*10^9)

Q = 3.3*10^-16 C = Charge in the fish

Part B.

Now charge is given by:

Q = n*e

n = number of electrons = ?

e = charge on a single electron = 1.6*10^-19 C

So,

n = Q/e = 3.3*10^-16/(1.6*10^-19)

n = 2062 electrons

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