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

Answer:

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).

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 = $\frac{kq_{1} q_{2} }{d^{2} }$

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 = $\frac{9.0*10^{9}*1.6*10^{-19}*1.6*10^{-19} }{(4.2*10^{-11}) ^{2} }$

  = $\frac{2.304*10^{-28} }{1.764*10^{-21} }$

  = 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.