Question

At what separation should two equal charges, $1 C$ each be placed so that the force between them equals the weight of a $50 kg$ person.

Answer 1

The two charges have the same magnitude, and the distance between the charges is not mentioned. It is said that the force required is equal to the weight of a 50kg person. Weight is nothing but the gravitational pull with which earth pulls all the bodies towards itself. So, we can find the required distance.

Complete step by step answer:
Mass of the body= 50 kg
Gravitational force, $F=mg$
$\Rightarrow F=50\times 9.8 \\\ \therefore F=490N$
From Coulomb’s law $F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{x}^{2}}}$, where $k$ is a constant which depends on the medium where the two charges are placed and $x$ is the distance between the two charges.
$\Rightarrow F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{x}^{2}}}=490N \\\ \Rightarrow \dfrac{k{{q}_{1}}{{q}_{2}}}{{{x}^{2}}}=490 \\\ \Rightarrow \dfrac{9\times {{10}^{9}}\times 1\times 1}{{{x}^{2}}}=490 \\\ \Rightarrow {{x}^{2}}=\dfrac{9\times {{10}^{9}}}{490} \\\ \Rightarrow x=0.42\times {{10}^{4}}m \\\ \therefore x=4200m$

So, the distance between the two charges comes out to be 4200 m.

Additional Information:
If a body is allowed to fall freely from a certain height then it falls under the influence of gravity and the acceleration of the body is g whose value is 9.8 $m/{{s}^{2}}$. Also, for making calculations we take the value of g = 10 instead of 9.8.

Note: The constant of proportionality $k$ that we have used while calculating is $\dfrac{1}{4\pi {{\varepsilon }_{0}}}$ where ${{\varepsilon }_{0}}$ is called permittivity of free space. If the medium is other than vacuum then we use $\varepsilon$. While doing such problems we have to keep in mind that all the units to be taken in standard SI units of the system. The universal law of gravitation holds everywhere. It is due to the gravitational force of the earth which holds all the bodies.