Question

An electron and a proton have charges of an equal magnitude but opposite sign of 1.60 x 10^-19 C. If the electron and proton in a hydrogen atom are separated by a distance of 4.20 x10^-11 m, what are the magnitude and direction of the electrostatic force exerted on the electron by the proton?

Answer 1

i. F = 1.3 x
`10^(-7)` N

ii. The direction of the force of attraction exerted by the proton on the electron is towards the itself (i.e a pull).

Step-by-step explanation:

Since the given charges are opposite, then the force of attraction is experienced. The force of attraction between the two charges can be determined by:

F =
`(kq_(1) q_(2) )/(d^(2) )`

where F is the force, k is the constant,
`q_(1)` is the charge of the electron,
`q_(2)` is the charge on the proton, and d is the distance between them.

So that; k = 9.0 x
`10^(9)` N
`m^(2)`
`C^(-2)` ,
`q_(1)` = 1.6 x
`10^(-19)` C,
`q_(2)` = 1.6 x

Thus,

F =
`(9.0*10^(9)*1.6*10^(-19)*1.6*10^(-19) )/((4.2*10^(-11)) ^(2) )`

=
`(2.304*10^(-28) )/(1.764*10^(-21) )`

= 1.3061 x
`10^(-7)`

F = 1.3 x
`10^(-7)` N

The force between the charges is 1.3 x
`10^(-7)` N.

ii. The direction of the force of attraction exerted by the proton on the electron is towards the itself.