Question

Calculate the coulombs force between two protons , if they are 4.0×10(-15) meters away from each other. (Where, charge of proton is 1.6×10(-19) coulomb.)

Answer 1

Answer:

14.4 N

Explanation:

Force for static electricity is -

$\dfrac{1}{4\pi\epsilon} \dfrac{q_1.q_2}{r^2}$

Here,

r = distance between them = $4.0 \times 10^{-15}$

$q_1 = q_2$ = charge on particles, here, both are same. So

$q_1 = q_2 = 1.6\times10^{-19}$

We know, $\dfrac{1}{4\pi\epsilon}=9 \times 10^9 [tex] </p><p>Hence, </p><p>[tex]Force=9 \times 10^9 \times \frac{1.6\times10^{-19}\times1.6\times10^{-19}}{(4.0\times10^{-15})^2}\\\\\\Force=9 \times 10^9 \times \frac{1.6\times10^{-19}\times1.6\times10^{-19}}{(16\times10^{-30}}\\\\\\ Force = 9\times \dfrac{1.6\times1.6}{16}\times10^{-19-19+30+9}\\\\\\ Force = 1.44 \times 10^{1}$

Force = 14.4 N