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

a)
`E=2.88*10^(-13)N/C`

b)
`E=1.44*10^(-13)N/C`

c)
`F=4.61*10^(-32)N`

Step-by-step 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`