Respuesta :
The magnitude of the electric force that an additional proton exerts on one proton is [tex]F_E = 57.6 N[/tex]
By Coulomb’s law,
[tex]$F_E=k \frac{q_1 q_2}{r^2}$[/tex] ----------------- equation(1)
We know that, Coulomb’s constant, [tex]k=9 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2 / \mathrm{C}^2[/tex].
Charge of the proton, [tex]q=1.6 \times 10^{-19} C[/tex]
Assume the distance between protons as, [tex]r=1.0 \mathrm{fm}=2.0 \times 10^{-15} \mathrm{~m}[/tex].
Substitute all the values in the equation(1),
[tex]\begin{aligned}F_E &=9 \times 10^9 \frac{\left(1.6 \times 10^{-19}\right)^2}{\left(2.0 \times 10^{-15}\right)^2} \\F_E &=9 \times 10^9 \frac{\left(2.56 \times 10^{-38}\right)}{\left(4.0 \times 10^{-30}\right)}\end{aligned}[/tex]
Magnitude of the Electric force is, [tex]F_E = 57.6 N[/tex]
Explain Coulomb’s law?
Coulomb's law is an experimental physical principle that measures the force exerted between two electrically charged particles that are stationary. Conventionally, the electric force between two charged objects at rest is referred to as the Coulomb force or electrostatic force.
Where, [tex]$F_E=k \frac{q_1 q_2}{r^2}$[/tex]
[tex]F =[/tex] Electric Force
[tex]k =[/tex]Coulomb's constant
[tex]q1,q2[/tex] = Charges
[tex]r =[/tex] Distance of separation
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