What is the magnitude of the force between an ion of charge 5e and an ion of charge -1e if their separation is 0.62 nm?

Question

What is the magnitude of the force between an ion of charge 5e and an ion of charge -1e if their separation is 0.62 nm?

F magnitude = ???

Is this force attractive or repulsive?

I got -1.17e+29 N as my answer which is incorrect but when indicating that the force is attractive that is correct. I used coulomb's law to solve for F and converted any units that were not in standard form (such as nm to m).

F = ((K (q1)(q2))/(r^2) = ((9e9Nm^2/C^2)((5*-1))/((6.2e-10)^2))= -1.17e+29 N. Not sure where I went wrong...

Answer 1

Hi Bri S.!

Your solution appears to take the elementary charge as equal to 1C. However, the actual value of the elementary charge can be approximated as:

e = 1.602x10^-19 C

To be fair, you can probably round to the tenth place for this question.

Plugging in our value for e, we find:

F_q = k|q₁||q₂|/r²

<=> (8.99x10⁹ N*m²*C^-2)|(5*1.6x10^-19C)||(-1.6x10^-19C)|/(6.2x10^-10 m)²

F_q = 2.99x10^-9 N

This represents the magnitude of our electrostatic force, however, it does not tell us about the attractive or repulsive nature of this interaction because we used the absolute value bars in our calculation. So, we must turn back to our question. The information tells us we have two charges involved, one with a charge equal to 5e and one equal to negative e. Because these charges are of opposite sign, we can deduce that the nature of the interaction is attractive.

Please let me know if that works out better for you!

Cheers