Question

An electron that is initially 52 cm away from a proton is displaced to another point. If the change in the electrical potential energy as a result of this movement is 2.4 × 10−28 J, what is the final distance between the electron and the proton? The Coulomb constant is 8.98755 × 109 N · m2 /C 2 , the acceleration due to gravity is 9.8 m/s 2 and the fundamental charge is 1.602 × 10−19 C . Answer in units of m.

Answer 1

Step-by-step explanation:

The given data is as follows.

Change in Potential Energy =
`2.4 * 10^(-28) J`

Formula for change in potential energy is as follows

Change in Potential Energy =
`charge * {\text{potential difference}}`

Hence, the potential difference between the two points will be as follows.

V =
`(2.4 * 10^(-28) J)/(-1.602 * 10^(-19))`

=
`-1.5 * 10^(-9)` volt

Therefore, the potential due to two charge initially is as follows.

V =
`V_(f) - V_(i)`

=
`-1.5 * 10^(-9)` volt

Hence,
`-1.5 * 10^(-9)` volt =
`9 * 10^(9) * 1.602 * 10^(-19)[(1)/(r) - (1)/(0.52)]`

r = 0.493 m = 49.3 cm (as 1 m = 100 cm)

Thus, we can conclude that the final distance between the electron and proton is 49.3 cm.