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# How far apart the two electrons are, if the force between them equals the weight of an electron? What in the case of protons?

Last updated date: 11th Aug 2024
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Hint: Before we approach the solution, it is important to understand the nature of force acting among the electrons and protons. Since they are charged bodies, there is an electrostatic force acting between them, which is given by Coulomb's law. Also, weight is equal to the mass times the acceleration due to gravity.

The problem has two parts: First part is if the charge is electron and second part is if the charge is a proton.

Part A: Electrons
Consider two electrons placed at a distance r. The electrostatic force acting between them is given by Coulomb's law which states the force between the two charges is directly proportional to the product of the charges and inversely proportional to the distance between the charges.
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
where ${\varepsilon _0}$ = absolute permittivity of free space.
The charge on an electron is equal to –
$e = 1. 602 \times {10^{ - 19}}C$
The weight of the electron is equal to the product of mass and acceleration due to gravity.
$W = mg$
Mass of electron, $m = 9. 1 \times {10^{ - 31}}kg$
Acceleration due to gravity, $g = 10m{s^{ - 2}}$
Weight, $W = 9. 1 \times {10^{ - 31}} \times 10 = 9. 1 \times {10^{ - 30}}N$
Given that the force acting between the electrons is equal to the weight of an electron, we have –
$F = W$
$\Rightarrow \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}} = W$
Substituting, we have –
$9 \times {10^9} \times \dfrac{{{{\left( {1. 6 \times {{10}^{ - 19}}} \right)}^2}}}{{{r^2}}} = 9. 1 \times {10^{ - 30}}$
$\Rightarrow {r^2} = \dfrac{{{{\left( {1. 6 \times {{10}^{ - 19}}} \right)}^2} \times 9 \times {{10}^9}}}{{9. 1 \times {{10}^{ - 30}}}}$
$\Rightarrow {r^2} = \dfrac{{2. 56 \times {{10}^{ - 38}} \times 9 \times {{10}^9}}}{{9. 1 \times {{10}^{ - 30}}}}$
$\Rightarrow {r^2} = \dfrac{{2. 56 \times 9}}{{9. 1}} \times {10^{ - 38 + 9 + 30}}$
$\Rightarrow {r^2} = 2. 5318 \times 10 = 25. 318$
$\therefore r = \sqrt {25. 318} = 5. 031m$
Thus, the distance of separation is equal to 5.03 m

Part B: Protons
Consider two protons placed at a distance r. The electrostatic force acting between them is given by Coulomb's law which states the force between the two charges is directly proportional to the product of the charges and inversely proportional to the distance between the charges.
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
where ${\varepsilon _0}$ = absolute permittivity of free space.
The charge on a proton is equal to –
$e = 1. 602 \times {10^{ - 19}}C$
The weight of the proton is equal to the product of mass and acceleration due to gravity.
$W = mg$
Mass of electron, $m = 1. 67 \times {10^{ - 27}}kg$
Acceleration due to gravity, $g = 10m{s^{ - 2}}$
Weight, $W = 1. 67 \times {10^{ - 27}} \times 10 = 1. 67 \times {10^{ - 26}}N$
Given that the force acting between the protons is equal to the weight of a proton, we have
$F = W$
$\Rightarrow \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}} = W$
Substituting, we have –
$9 \times {10^9} \times \dfrac{{{{\left( {1. 6 \times {{10}^{ - 19}}} \right)}^2}}}{{{r^2}}} = 1. 67 \times {10^{ - 26}}$
$\Rightarrow {r^2} = \dfrac{{{{\left( {1. 6 \times {{10}^{ - 19}}} \right)}^2} \times 9 \times {{10}^9}}}{{1. 67 \times {{10}^{ - 26}}}}$
$\Rightarrow {r^2} = \dfrac{{2. 56 \times {{10}^{ - 38}} \times 9 \times {{10}^9}}}{{1. 67 \times {{10}^{ - 26}}}}$
$\Rightarrow {r^2} = \dfrac{{2. 56 \times 9}}{{1. 67}} \times {10^{ - 38 + 9 + 26}}$
$\Rightarrow {r^2} = 13. 796 \times {10^{ - 3}} = 0. 0137$
$\therefore r = \sqrt {0. 0137} = 0. 117m$

Hence, the distance of separation is equal to 0.117m.

Note: The proton and electron have the same charge, known as the fundamental charge $e = 1. 602 \times {10^{ - 19}}C$, but they are opposite in sign. While proton has charge equal to $+ e$, the charge of electrons is $- e$. However, in this question, only the magnitude of the charge is considered in the calculations and not the sign.