Question

Two particles with charges +6e and -6e are initially very far apart (effectively an infinite distance apart). They are then fixed at positions that are 5.61 x 10^-12 m apart. What is EPEfinal - EPEinitial, which is the change in the electric potential energy?

Answer 1

`\rm EPE_(final)-EPE_(initial)=-1.478* 10^(-15)\ J.`

Step-by-step explanation:

Given charges are:

`\rm q_1 = +6e.\\q_2 = -6e.`

The electric potential energy of a charge due to the electric field of another charge is given by

`\rm EPE=(kq_1q_2)/(r).`

where,

• k = Coulomb's constant, having value =
  `\rm 9* 10^9\ Nm^2/C^2.`
• r = distance between the charges.

When the charges are infinite distance apart,
`\rm r = \infty`,

`\rm EPE_(initial) = (kq_1q_2)/(r)=0\ J.`

When the charges are
`\rm 5.61* 10^(-12)\ m` apart,
`\rm r=5.61* 10^(-12)\ m`,

`\rm EPE_(final)=(kq_1q_2)/(r)\\=((9* 10^9)* (+6e)* (-6e))/(5.61* 10^(-12))\\=-5.775\ e^2* 10^(22).`

Here, e is the charge on one electron, such that,
`\rm e = -1.6* 10^(-19)\ C`.

Therefore,

`\rm EPE_(final)=-5.775* (-1.6* 10^(-19))^2* 10^(22) = -1.478* 10^(-15)\ J.`

Thus,

`\rm EPE_(final)-EPE_(initial)=-1.478* 10^(-15)-0=-1.478* 10^(-15)\ J.`