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Two electrons are (20 mm) apart at closest approach. What is the magnitude of the maximum electric force that they exert on each other? (e = 1.60 × 101° C, k = 1/47ɛ, = 9.0 10° N · m²/C³). -19 %3D

Respuesta :

Answer:

Answer is 5.76×10^-25.

Explanation:

Given values in the question are:

q = 1.6 x 10⁻¹⁹ C

k = 9 x 10⁹ N.m²/C²

Hence two points is 20mm apart so the radius becomes

R=0.02.

The formula is

F = kq^2/r^2

F= 9 x 10⁹× (1.6 x 10⁻¹⁹)^2 /(0.02)^2

F=5.76×10^-25.

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