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Two equal and like charges when placed 50m apart experience a repulsive force of 144N. The magnitude of charge in micro-coulomb will be:

Last updated date: 22nd Jun 2024
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Hint: Just like any other numerical problem, firstly, you could note down all the given values. Then you could recall Coulomb’s law. After recalling, you could substitute all the given values into the expression for Coulomb’s law. Then, you could rearrange this expression to get the required magnitude of charge.
Formula used:
Coulomb’s law,

Complete answer:
In the question, we are given two charges that are identical in both sign and magnitude that are separated by 50meters. Since these charges are of the same sign, quite obviously the force experienced by them would be repulsive and the magnitude of this repulsive force is given as 144N. Using all this information, we are supposed to find the magnitude of this charge in units of micro-coulombs.
In order to answer this question, let us recall Coulomb’s law. The statement of this law goes like this. The force between two charges separated by a certain distance is found to be proportional to the product of the charges and inversely proportional to the distance that separates them. Mathematically,
Here, k acts as the proportionality constant.
In the given question,
Substituting all the given values,
$144=9\times {{10}^{9}}\times \dfrac{{{q}^{2}}}{{{50}^{2}}}$
$\Rightarrow {{q}^{2}}=\dfrac{144\times 2500}{9\times {{10}^{9}}}$
$\Rightarrow q=0.63245\times {{10}^{-2}}C$
$\therefore q=6324.5\mu C$
Therefore, we found the magnitude of the charge in microcoulombs to be$q=6324.5\mu C$.

You may have noticed the proportionality constant that we introduced in the coulomb’s law. This constant is known to have a value given by,
$k=\dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{C}^{-2}}{{m}^{2}}$
Though there is no mention of signs in the solution, it does have significance. Taking the product will cancel the sign out when they are like charges.