What is the electric potential energy of a system that consists of two protons 2.1×10−15 m apart-as might occur inside a typical nucleus? Express your answer using two significant figures.

Respuesta :

Answer:

[tex]1.1\times 10^{-12}\ J[/tex]

Explanation:

k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]

q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

r = Distance of charges = [tex]2.1\times 10^{-15}\ m[/tex]

Electric potential energy is given by

[tex]V=\dfrac{kqq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^9\times (1.6\times 10^{-19})^2}{2.1\times 10^{-15}}\\\Rightarrow V=1.0959238095\times 10^{-13}\ V[/tex]

The electric potential energy of the system is [tex]1.1\times 10^{-13}\ J[/tex]

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