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Question: Two point charges with charges \(3\) micro coulombs and \(4\) micro coulombs are separated by \(2\;c...

Two point charges with charges 33 micro coulombs and 44 micro coulombs are separated by 2  cm2\;cm. The value of the force between them?
(A) 600  N600\;N
(B) 300  N300\;N
(C) 540  N540\;N
(D) 270  N270\;N
(E) 400  N400\;N

Explanation

Solution

We have two point charges having 33 micro coulomb and 44 micro coulomb each. The distance between both the charges is given by 2  cm2\;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=14πε0Q1Q2r2F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{Q_1}{Q_2}}}{{{r^2}}}
where, FF stands for the force between the two charges, ε0{\varepsilon _0} is the permittivity of free space, Q1{Q_1} and Q2{Q_2} are the two charges and rr 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, Q1=3μC{Q_1} = 3\mu C
Converting into Coulomb by multiplying with 106{10^{ - 6}}, Q1=3×106C{Q_1} = 3 \times {10^{ - 6}}C
The value of the second charge is given as, Q2=4μC{Q_2} = 4\mu C
Converting into Coulomb by multiplying with 106{10^{ - 6}}, Q2=4×106{Q_2} = 4 \times {10^{ - 6}}
The distance between both charges is given as, r=2cm=0.02mr = 2cm = 0.02m
Coulomb’s law is given by,
F=14πε0Q1Q2r2F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{Q_1}{Q_2}}}{{{r^2}}}
Substituting the values within the above equation, we get
F=14πε0(3×106)×(4×106)(0.02)2F = \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 14πε0=9×109Nm2C2\dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}
Substituting within the above equation, we get
F=9×109×(3×106)×(4×106)(0.02)=270NF = \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  N270\;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.