Answer
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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): $270\;N$
Additional Information:
The magnitude of coulomb charges will depend on three factors that are the distance between the charges, the number of charges, and the nature of the media between the charges. Positive charges are attractive in nature meanwhile negative charges are repulsive in nature.
Note: Coulomb’s law states that the force of attraction or repulsion between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them. Like charges will have a repulsive force between them and unlike charges will have an attractive force between them.
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): $270\;N$
Additional Information:
The magnitude of coulomb charges will depend on three factors that are the distance between the charges, the number of charges, and the nature of the media between the charges. Positive charges are attractive in nature meanwhile negative charges are repulsive in nature.
Note: Coulomb’s law states that the force of attraction or repulsion between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them. Like charges will have a repulsive force between them and unlike charges will have an attractive force between them.
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