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]