Question

Consider an electron that is 100 m from an alpha particle ( = 3.2 x 10-19 C). (Enter the magnitudes.) (a) What is the electric field (in N/C) due to the alpha particle at the location of the electron? N/C (b) What is the electric field (in N/C) due to the electron at the location of the alpha particle? N/C (c) What is the electric force (in N) on the alpha particle? On the electron? electric force on alpha particle electric force on electron

Answer 1

Answer:

a)$E=2.88*10^{-13}N/C$

b)$E=1.44*10^{-13}N/C$

c)$F=4.61*10^{-32}N$

Explanation:

The definition of a electric field produced by a point charge is:

$E=k*q/r^2$

a)Electric Field due to the alpha particle:

$E=k*q_{alpha}/r^2=9*10^9*3.2*10^{-19}/(100)^2=2.88*10^{-13}N/C$

b)Electric Field  due to the electron:

$E=k*q_{electron}/r^2=9*10^9*1.6*10^{-19}/(100})^2=1.44*10^{-13}N/C$

c)Electric Force on the alpha particle, on the electron:

The alpha particle and electron feel the same force magnitude but with opposite direction:

$F=k*q_{electron}*q_{alpha}/r^2=9*10^9*1.6*10^{-19}*3.2*10^{-19}/(100)^2=4.61*10^{-32}N$