Question

A proton is located at the origin, and a second proton is located on the x-axis at x = 5.22 fm (1 fm = 10-15 m).

(a) Calculate the electric potential energy associated with this configuration.(b) An alpha particle (charge = 2e, mass = 6.64 10-27 kg) is now placed at (x, y) = (2.61, 2.61) fm. Calculate the electric potential energy associated with this configuration.

Answer 1

(a)
`U_e=4.41*10^(-14)J`

(b)
`U_e=2.94*10^(-13) J`

Explanation:

(a)The formula for the electric potential energy is:

`U_e=k_e(q_1q_2)/(r)`

Where
`k_e` is the Coulomb's constant and is equal to
`9*10^9Nm^2/C^2`,
`q_1` and
`q_2`are the charges and
`r` is the distance between them. In this case
`r=5.22*10^(-15)m`,
`q1=q2=1.60*10^(-19)C`.

`U_e=k_e(q_1q_2)/(r)=(9*10^9)((1.60*10^(-19))(1.60*10^(-19)))/(5.22*10^(-15)) \\U_e=(9*10^9)(25.6*10^(-39))/(5.22*10^(-15))=(9*10^9)(4.90*10^(-24))\\U_e=4.41*10^(-14)J`

(b) In systems with three charges, the electric potential energy is the sum of the contribution made by each pair of the charges, i.e. the electric potential energy of the charges
`q_1` and
`q_2` plus the electric potential energy of the charges
`q_2` and
`q_3` plus the electric potential energy of the charges
`q_3` and
`q_1`.

`U_e=k_e(q_1q_2)/(r_(12))+k_e(q_2q_3)/(r_(23))+k_e(q_3q_1)/(r_(31))`

`q_1=q_2=1.60*10^(-19)C`

`q3=2e=2(1.60*10^(-19))=3.2*10^(-19)C`

Since
`2.61*10^(-15)m` is half of
`5.22*10^(-15)m`, the charge
`q_3` is at the same distance from
`q_1` as from
`q_2` and is the hypotenuse of the triangle rectangle with two equal sides of
`2.61*10^(-15)m`.

`r_(23)=r_(31)=\sqrt{(2.61*10^(-15))^2+(2.61*10^(-15))^2} =3.69*10^(-15)m`

`r_(12)=5.22*10^(-15)m`

`U_e=k_e(q_1q_2)/(r_(12))+k_e(q_2q_3)/(r_(23))+k_e(q_3q_1)/(r_(31))`

`U_e=k_e((1.60*10^(-19))(1.60*10^(-19)))/(5.22*10^(-15))+k_e((1.60*10^(-19))(3.20*10^(-19)))/(3.69*10^(-15))+k_e((3.20*10^(-19))(1.60*10^(-19)))/(3.69*10^(-15))`

`U_e=k_e(2.56*10^(-38))/(5.22*10^(-15))+k_e(5.12*10^(-38))/(3.69*10^(-15))+k_e(5.12*10^(-38))/(3.69*10^(-15))`

`U_e=k_e4.90*10^(-24)+k_e1.39*10^(-23)+k_e1.39*10^(-23)`

`U_e=(9*10^9)(4.90*10^(-24))+(9*10^9)(1.39*10^(-23))+(9*10^9)(1.39*10^(-23))`

`U_e=4.41*10^(-14)+1.25*10^(-13)+1.25*10^(-13)`

`U_e=2.94*10^(-13) J`