Question

Two protons are 1m apart, how much force is in their repulsion?

Answer 1

use coulomb's law?

Answer 2

This one?\[\left| F \right|=k_e\frac{\left|q_1q_2\right|}{r^2}\]

Answer 3

yep

Answer 4

\[\left| F \right|=8.99\times 10^9N\cdot m^2\cdot C^{-2}\frac{\left|(1.60217656535\times 10^{−19}C)(1.60217656535\times 10^{−19}C)\right|}{1m^2}\]
\[\left| F \right|=8.99\times 10^9N\cdot\cancel{m^2}\cdot\cancel{C^{-2}}\frac{\left|(1.60217656535\times 10^{−19}\cancel{C})(1.60217656535\times 10^{−19}\cancel{C})\right|}{\cancel{1m^2}}\]
\[F=2.3\times 10^{-29}N\]So, not very much then.

Answer 5

sorry, i was double checking that "r" wasn't radius. But yeah, that's like nothing.

Answer 6

Thanks

Answer 7

no problem