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 ?N/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 69 cm? How many electrons would be required to create that charge?

Answer 1

(a) E = k*q/(distance^2)
=>3 *10^-6 = (9*10^9)*q/(55*10^-2)^2
=>q = 1.008 *10^-16 C

(b) n = q/e
=>n = (1.008 *10^-16)/(1.6*10^-19) = 630 electrons

Answer 2

a)kq/r²=3*10^-6

q=2.187*10^-16C

b)No. of electrons=2.187*10^-16C/1.6*10^-19C=1367

Answer 3

| Answer: given dala Electric field E= 3x10C The electric due to a point ,r=0.69m charge at a distance Es Kuya 3x10 6 =9x109xq (0.69) 2=3x10-6x0.69 9x109 ::9=15.87 x10170 nowe The number of elections are 15.87 X1017 n= T. 6x 10-19 .: n=991.88