Answer:
[tex]dU=-4.36*10^{-18}J[/tex]
Explanation:
From the question we are told that:
Average distance [tex]d_{avg} =5.29*10^{-11}m[/tex]
Generally the equation for change in electric potential energy is mathematically given by
[tex]dU=u_f-U_1[/tex]
Where
U_1=0 Because of initial lengthy distance apart
And
[tex]U_f=\frac{kq_eq_p}{d}[/tex]
[tex]U_f=\frac{9*10^9*1.6*10^{-19}*-1.6*10^{-19}}{5.29*10^{-11}}[/tex]
[tex]U_f=-4.36*10^{-18}J[/tex]
Therefore
[tex]dU=u_f-U_1[/tex]
[tex]dU=-4.36*10^{-18}J-0[/tex]
[tex]dU=-4.36*10^{-18}J[/tex]