Use Coulomb's Law to calculate the number of excess electrons

[dwach]

Homework Statement

Two negatively charged objects repel each other with a measured force of 6.3 N when they are 0.5 cm apart. If the excess charge on one of the objects is caused by 8.3 x 10^22 extra electrons, use Coulomb's Law to calculate the number of excess electrons on the second object.

Homework Equations

F = k*(q1)*(q2) / r^2

1e = 1.602 * 10^-19 c
1c = 6.242 * 10^18 e
k = 9.0 * 10^9 N*m^2/c^2

The Attempt at a Solution

I calculated the excess charge to be:

(8.3*10^22 e) * (1.602 *10^-19) = 1.3 * 10^4 c

I used this as q2 in the F = k*(q1)*(q2) / r^2 equation and rearranged to find that:

q1 = (F * r^2) / k*q2
q1 = (6.3N * 0.005m^2) / (9.0*10^9N*m^2/c^2)*(1.3*10^4c)
q1 = 1.3 * 10^-18

Since 1c = 6.242*10^18e, to find number of excess electrons i multiplied q1 by 6.242*10^18 = 8.1146 eSo is the number of excess electrons on the second object 8.1 ?

 

[Delphi51]

Welcome to PF, dwach!
Your calc looks good, and thank you for taking time to write out the method so clearly. I don't know what accuracy you need; I'm used to 3 digits so I keep 4 in intermediate steps and I got 1.318 x 10^-18 for the charge, which is 8.24 electrons.

 

[dwach]

Thank you for the quick response Delphi51. In this question I need 2 digit accuracy so would you suggest that I carry 3 digits through the intermediate steps?

 

[Delphi51]

Yes, you must carry 3 digits to get 2 at the end. In practise, you probably will just keep all the digits your calculator produces for the first calc and only round when you write down the final answer.