Answer:
the electrostatic force between the two protons is 5.775 x 10⁻²⁹ N.
Explanation:
Given;
charge of protons, q = 1.602 x 10⁻¹⁹ C
distance between the two charges, r = 2.0 m
The electrostatic force between the two protons is calculated as;
[tex]F = \frac{kq^2}{r^2}[/tex]
where;
k is coulomb's constant, = 9 x 10⁹ Nm²/C²
[tex]F = \frac{(9\times 10^9)(1.602 \times 10^{-19})^2}{(2)^2} \\\\F = 5.775 \times 10^{-29} \ N[/tex]
Therefore, the electrostatic force between the two protons is 5.775 x 10⁻²⁹ N.
The force between the two protons separated by the distance of 2 m will be [tex]F=5.775\times 10^{-29}\ N[/tex]
The coulomb's force is defined as when two charged particles are separated by the distance r and having charge q then the force will be directly proportional to the product of the charges and inverse to the distance between them.
The coulomb's force will be given by
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
Here it is given that
Charge on the proton=[tex]q=1.602\times 10^{-19}\ C[/tex]
The distance between protons = 2 m
The value of constant [tex]k=9\times 10^{-29}\ \dfrac{Nm^2}{C^2}[/tex]
Putting the values in the formula
[tex]F=\dfrac{(9\times 10^9) (1.602\timeds 10^{-19}^)^2}{2^2}[/tex]
[tex]F=5.775\times 10^{-29}\ N[/tex]
Hence the force between the two protons separated by the distance of 2 m will be [tex]F=5.775\times 10^{-29}\ N[/tex]
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