A platypus foraging for prey can detect an electric field as small as 0.002 N/C. Part A To give an idea of sensitivity of the platypus’s electric sense, how far from a 50nC point charge does the field have this magnitude

Answer:

474.34 m

Explanation:

From the question,

E = kq/r²…………….. Equation 1

Where E = Electric Field, k = coulomb’s constant, q = Charge, r = distance.

Answer:474.34 mExplanation:From the question,

E = kq/r²…………….. Equation 1

Where E = Electric Field, k = coulomb’s constant, q = Charge, r = distance.

Make r the subject of the equation

r = √(kq/E)………… Equation 2

Given: q = 50 nC = 50×10⁻⁹ C, E = 0.002 N/C

Constant: k = 9×10⁹ Nm²/C².

Substitute these values into equation 2

r = √(50×10⁻⁹×9×10⁹/0.002)

r = √(450/0.002)

r = √(225000)

r = 474.34 m