Question

An electron is held up against the force of gravity by the attraction of a fixed proton some distance above it. How far above the electron is the proton

Answer 1

5.08 m

Explanation:

The weight of the electron is being counteracted by the attractive electrostatic force exerted by the proton above it. We can write the force equation as follows:

$m_eg = k_e\dfrac{Q_pQ_e}{r^2}$

where the Q's are the charges of the proton and electron, r is the distance between the particles, g is the acceleration due to gravity, $m_e$ is the mass of the electrons and $k_e$ is the Coulomb constant. So solving for r, we get

$r^2 = k_e\dfrac{Q_pQ_e}{m_eg}$

Taking the square root of r^2, we then get the distance as

$r = \sqrt{k_e\dfrac{Q_pQ_e}{m_eg}}$

The values are given as follows:

$m_e = 9.11×10^{-31}\:\text{kg}$

$g = 9.8\:\text{m/s}^2$

$Q_p = Q_e = 1.60×10^{-19}\:\text{C}$

$k_e = 8.99×10^9\:\text{N-m}^2\text{/C}^2$

Putting in all of these values in our equation for r,

$r = \sqrt{\dfrac{(8.99×10^9\:\text{N-m}^2\text{/C}^2)(1.60×10^{-19}\:\text{C})^2}{(9.11×10^{-31}\:\text{kg})(9.8\:\text{m/s}^2)}}$

$\:\:\:\:\:= 5.08\:\text{m}$