Order of magnitude problem involving the force of repulsion between two persons

[StrawHat]

Homework Statement

Homework Equations

$\vec{F}$ = k$\stackrel{q₁q₂}{r^{2}}$

The Attempt at a Solution

6.022e23*(0.505) = 3.041e23C <-- electrons
6.022e23*(0.495) = 2.981e23C <-- protons
3.041e23 - 2.981e23 = 6e21C <-- the difference between the two charges
$\vec{F}$$_{e}$ = (9e9Nm$^{2}$/C$^{2}$)(6e21C)$^{2}$ / 2.25m$^{2}$
$\vec{F}$$_{e}$ = 1.44e53 N

 

[CAF123]

Given that the force of repulsion is similar to the gravitational force exerted on an object with a mass the size of earth, you should expect F ~ 10²⁵N.

There is an identical thread here:
https://www.physicsforums.com/showthread.php?t=431698

 

[StrawHat]

Given that the force of repulsion is similar to the gravitational force exerted on an object with a mass the size of earth, you should expect F ~ 10²⁵N.

There is an identical thread here:
https://www.physicsforums.com/showthread.php?t=431698

The answer gives me this error: "Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error."

 

[nasu]

What is the answer that you put in the box?

 

[StrawHat]

What is the answer that you put in the box?

I put in F~10 25N.

 

[voko]

How do you get the number of protons?

 

[nasu]

I put in F~10 25N.

I would consider trying to enter 26 in the box. Depending on how you estimate, you may get 10^26. And the "weight" of the Earth is also of the order 10^26 (6x10^25).