Effect of electrical potential energy in the binding energy of an nucleus?

The binding energy of a nucleus with $$Z$$ protons and $$N=A-Z$$ neutrons is given by $$B=(Zm_p+Nm_n-m_{nucleus})c^2$$
In the above equation we subtract the rest mass of nucleus from the rest mass of nucleons in free states but we are not accounting for the electrical potential energy between protons. Is it because it is negligible or is it included in $$m_{nucleus}$$ or is it any other reason?

Well, It actually already included. The rest mass of nuclei $$m_\text{nuc}c^2$$ include all sorts of energies. When we subtract the individual masses with this mass, we get amount of mass due these interactions.
$$B(N,Z)=aA-bA^{2/3}-s\frac{(N-Z)^2}{A}-\frac{dZ^2}{A^{1/3}}-\frac{\delta }{\sqrt{A}}$$ The fourth term account for coulomb interaction.