Question

Consider the electrostatic force between two particles of equal mass m = 32 × 10^-26 kg and carrying different charges q1 = 17e and q2 = 28e, where e denotes the magnitude of the charge on an electron. At what distance, in meters, is the electrostatic force between these two particles equal to the weight of one of them?

Answer 1

Answer: 0.187 m

Explanation: In order to explain this problem we have to consider the Coulomb and gravitational force, which are given by;

Electric force

Fe=(k*q1*q2)/d^2 where k is a constant equal to 9*10^9 N/C^2*m^2 and d is the distanvce between the charges.

gravitational force

Fg=me*g  

equaling both forces,  we have:

(k*q1*q2)/d^2 = me*g  

d^2= (k*q1*q2)/me*g = 9*10^9*17*28*(1.6*10^-19)^2/(32*10^-26*9.8)=

=0.187 m

Answer 2

The distance should be 187 m

Calculation of the distance is:

Fe=(k*q1*q2)/d^2

Here k is a constant equal to 9*10^9 N/C^2*m^2

And d is the distance between the charges.

Now

Fg=me*g  

Now we have to equate both the equations

So,

(k*q1*q2)/d^2 = me*g  

d^2= (k*q1*q2)/me*g = 9*10^9*17*28*(1.6*10^-19)^2/(32*10^-26*9.8)=

=0.187 m

learn more about the electron here: https://brainly.com/question/11624072