An electron and a proton have charges of opposite sign, but the magnitude of their charges has the same value of 1.60 10-19 C. Suppose the electron and proton in a hydrogen atom are separated by a distance of 4.30 10-11 m. What are the magnitude and direction of the electrostatic force exerted on the electron by the proton?
clearly the force is attractive, pointing toward the proton.
F = kqq/r^2
= (8.99*10^9)(1.60*10^-19)^2/(4.30*10^-11)^2
= 1.24*10^-7 N
To find the magnitude and direction of the electrostatic force exerted on the electron by the proton, we can use Coulomb's law. Coulomb's law states that the magnitude of the electrostatic force between two charged particles is given by the equation:
F = k * (q1 * q2) / r^2
where F is the magnitude of the force, k is the electrostatic constant (8.99 x 10^9 N m^2/C^2), q1 and q2 are the charges of the particles, and r is the distance between the particles.
In this case, the charge of the electron (q1) is -1.60 x 10^-19 C, the charge of the proton (q2) is +1.60 x 10^-19 C, and the distance between them (r) is 4.30 x 10^-11 m. Substituting these values into Coulomb's law, we can calculate the magnitude of the force:
F = (8.99 x 10^9 N m^2/C^2) * (1.60 x 10^-19 C) * (1.60 x 10^-19 C) / (4.30 x 10^-11 m)^2
Performing the calculations, the magnitude of the force is approximately 8.21 x 10^-8 N.
The direction of the force is attractive since the electron and proton have charges of opposite sign. This means the electron will be attracted towards the proton.
Therefore, the magnitude of the electrostatic force exerted on the electron by the proton is 8.21 x 10^-8 N and the direction of the force is towards the proton.