Question

Calculate the magnitude of electrastatri force between a proton and an electron separated by a distance of 0.5 A°

Answer 1

Answer:

$9.26 \times 10^{-8}\ N$ is the magnitude of electrostatic force.

Explanation:

We are given that there is a proton and an electron.

Distance between them, r = 0.5 A°

To find:

The magnitude of electrostatic force between the two.

Formula for electrostatic force is given as:

$F_e = k \dfrac{q_1q_2}{r^{2} }$

Where $q_1, q_2$ are the amount of charge on the objects

r is the distance between the two objects.

k is proportionality constant

Let us have a look at the values of different variables:

$k = 9\times 10^9 Nm^2/C^2$

$q_1=q_2=1.6 \times 10^{-19} C$

$r = 0.5 \AA = 0.5 \times 10^{-10}\ m$

Putting the values to find $F_e$:

$F_e = (9 \times 10^9) \dfrac{(1.6 \times 10^{-19}) \times (1.6 \times 10^{-19})}{(0.5 \times 10^{-10})^{2} }\\\Rightarrow F_e = (9 \times 10^9) \dfrac{(2.56 \times 10^{-38})}{(0.25 \times 10^{-20}) }\\\Rightarrow F_e = (9 \times 2.56) \dfrac{( 10^{-9})}{(0.25) }\\\Rightarrow F_e = 9.26 \times 10^{-8}\ N$

So, the answer is:

$9.26 \times 10^{-8}\ N$ is the magnitude of electrostatic force.