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# Two point charges with charges $3$ micro coulombs and $4$ micro coulombs are separated by $2\;cm$. The value of the force between them?(A) $600\;N$(B) $300\;N$(C) $540\;N$(D) $270\;N$(E) $400\;N$

Last updated date: 17th Apr 2024
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Hint: We have two point charges having $3$ micro coulomb and $4$ micro coulomb each. The distance between both the charges is given by $2\;cm$. We have to find the force between them. This question is a direct application of Coulomb’s law and can be easily solved by applying coulomb’s law. The values are all given.
Formula used:
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{Q_1}{Q_2}}}{{{r^2}}}$
where, $F$ stands for the force between the two charges, ${\varepsilon _0}$ is the permittivity of free space, ${Q_1}$ and ${Q_2}$ are the two charges and $r$ stands for the distance between the two charges.

Complete step by step solution:
Both charges are separated by a distance.
The value of the first charge is given as, ${Q_1} = 3\mu C$
Converting into Coulomb by multiplying with ${10^{ - 6}}$, ${Q_1} = 3 \times {10^{ - 6}}C$
The value of the second charge is given as, ${Q_2} = 4\mu C$
Converting into Coulomb by multiplying with ${10^{ - 6}}$, ${Q_2} = 4 \times {10^{ - 6}}$
The distance between both charges is given as, $r = 2cm = 0.02m$
Coulomb’s law is given by,
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{Q_1}{Q_2}}}{{{r^2}}}$
Substituting the values within the above equation, we get
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{\left( {3 \times {{10}^{ - 6}}} \right) \times \left( {4 \times {{10}^{ - 6}}} \right)}}{{{{\left( {0.02} \right)}^2}}}$
The value of $\dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}$
Substituting within the above equation, we get
$F = \dfrac{{9 \times {{10}^9} \times \left( {3 \times {{10}^{ - 6}}} \right) \times \left( {4 \times {{10}^{ - 6}}} \right)}}{{\left( {0.02} \right)}} = 270N$

The answer is: Option (D): $270\;N$