Respuesta :
Answer:
[tex]\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}\ J.[/tex]
Explanation:
Given charges are:
[tex]\rm q_1 = +6e.\\q_2 = -6e.[/tex]
The electric potential energy of a charge due to the electric field of another charge is given by
[tex]\rm EPE=\dfrac{kq_1q_2}{r}.[/tex]
where,
- k = Coulomb's constant, having value = [tex]\rm 9\times 10^9\ Nm^2/C^2.[/tex]
- r = distance between the charges.
When the charges are infinite distance apart, [tex]\rm r = \infty[/tex],
[tex]\rm EPE_{initial} = \dfrac{kq_1q_2}{r}=0\ J.[/tex]
When the charges are [tex]\rm 5.61\times 10^{-12}\ m[/tex] apart, [tex]\rm r=5.61\times 10^{-12}\ m[/tex],
[tex]\rm EPE_{final}=\dfrac{kq_1q_2}{r}\\=\dfrac{(9\times 10^9)\times (+6e)\times (-6e)}{5.61\times 10^{-12}}\\=-5.775\ e^2\times 10^{22}.[/tex]
Here, e is the charge on one electron, such that, [tex]\rm e = -1.6\times 10^{-19}\ C[/tex].
Therefore,
[tex]\rm EPE_{final}=-5.775\times (-1.6\times 10^{-19})^2\times 10^{22} = -1.478\times 10^{-15}\ J.[/tex]
Thus,
[tex]\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}-0=-1.478\times 10^{-15}\ J.[/tex]
