Question

E

A proton and an electron in a hydrogen

atom are separated on the average by about

5 × 10^−11 m.

What is the magnitude of the electric field

set up by the proton at the location of the

electron? The value of the Coulomb constant

is 8.99 × 10^9 N · m^2/C^2

.

Answer in units of N/C.

Answer 1

Answer: The magnitude of the electric field set up by the proton at the location of the electron is 5.68 × 10^11 N/C.

Explanation: The electric field at a distance r from a point charge q is given by Coulomb’s law as:

E = kq/r^2

where k is Coulomb’s constant (8.99 × 10^9 N · m2/C2).

In this case, the distance between the proton and electron is r = 5 × 10^-11 m. Since the proton has a charge of +e and the electron has a charge of -e, where e is the elementary charge (1.602 × 10^-19 C), we have:

E = kq/r^2 = (8.99 × 10^9 N · m2/C2) × [(+e)(-e)]/(5 × 10^-11 m)^2 = 5.68 × 10^11 N/C.

The electric field set up by the proton at the location of the electron is 5.68 × 10^11 N/C.

Hope this helps, and have a great day! =)