Question

what is the approximate electrostatic force between two protons separated by a distance of 1.0x10^-6

Answer 1

Final Answer :
$2.3 \times {10}^{ - 16} \: N$

Steps :
1) Charge =q(1) = q(2) = 1.6 * 10^(-19) C
Distance, r = 1.0 * 10^(-6) m

By Coulomb Law,
F (elect) =Kq(1)q(2) / r^2

$= \frac{9 \times {10}^{9} \times 1.6 \times {10}^{ - 19} \times 1.6 \times {10}^{ - 19} }{ ({1.0 \times {10}^{ - 6} )}^{2} } \\ \\ = 9 \times 1.6 \times 1.6 \times {10}^{ - 16} \\ = 2.3 \times {10}^{ - 16} \: N$

Therefore, Electrostatic Force of attraction between two protons is 2.3 * 10^(-16) N.

Answer 2

Answer:

idk

Explanation:

need pts