Answer
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Hint-The coulomb’s law is in between the particles at atomic distance of\[{10^{ - 10}}m\]. The like charges repel each other hence the coulomb’s force between like charges will be the repulsive force. And the coulomb force between unlike charges will be the force of attraction.
Complete step by step answer:
Coulomb’s law:
The coulomb’s law states that the force of attraction or repulsion between two atoms is directly proportional to the product of the charges and inversely proportional to the square of the distance between them
(i) According to Coulomb’s law, the force of attraction or repulsion,
\[F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}\] -------- (1)
Where
\[k\]- The constant of proportionality.
\[{q_1},{q_2}\]- Charges in contact,
\[r\]- Distance between the charges
(ii) From the (1), the constant
\[ \Rightarrow k = \dfrac{{F{r^2}}}{{{q_1}{q_2}}}\]
Applying the SI units of the quantity in their respective places, therefore
\[ \Rightarrow k = \dfrac{{N{m^2}}}{{{C^2}}}\]
This is the unit of the constant of proportionality, Hence the correct option is D.
Additional information:
(i) Between the objects at larger distances, there is a gravitational force. Between the particles at smaller distances which is below the atomic distance, there is a coulomb force.
(ii) The Coulomb’s force is greater than the gravitational force as it is between the objects at larger distances. We know that the force is inversely proportional to the distance. Hence the distance increases and attraction decreases.
(iii) The constant of proportionality $K$ is also greater than the gravitational constant.
Note:(i) The Coulomb force is the strongest force as it is involved between the objects at smaller distances. The value of the constant of proportionality equals to \[9 \times {10^9}N{m^2}{C^{ - 2}}\].
(ii) The unit of charge is ‘coulomb’ which is denoted as ‘$C$’
(iii) The unit of distance is measured in metres ‘$m$’. And the force is measured in Newton ‘$N$’
Complete step by step answer:
Coulomb’s law:
The coulomb’s law states that the force of attraction or repulsion between two atoms is directly proportional to the product of the charges and inversely proportional to the square of the distance between them
(i) According to Coulomb’s law, the force of attraction or repulsion,
\[F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}\] -------- (1)
Where
\[k\]- The constant of proportionality.
\[{q_1},{q_2}\]- Charges in contact,
\[r\]- Distance between the charges
(ii) From the (1), the constant
\[ \Rightarrow k = \dfrac{{F{r^2}}}{{{q_1}{q_2}}}\]
Applying the SI units of the quantity in their respective places, therefore
\[ \Rightarrow k = \dfrac{{N{m^2}}}{{{C^2}}}\]
This is the unit of the constant of proportionality, Hence the correct option is D.
Additional information:
(i) Between the objects at larger distances, there is a gravitational force. Between the particles at smaller distances which is below the atomic distance, there is a coulomb force.
(ii) The Coulomb’s force is greater than the gravitational force as it is between the objects at larger distances. We know that the force is inversely proportional to the distance. Hence the distance increases and attraction decreases.
(iii) The constant of proportionality $K$ is also greater than the gravitational constant.
Note:(i) The Coulomb force is the strongest force as it is involved between the objects at smaller distances. The value of the constant of proportionality equals to \[9 \times {10^9}N{m^2}{C^{ - 2}}\].
(ii) The unit of charge is ‘coulomb’ which is denoted as ‘$C$’
(iii) The unit of distance is measured in metres ‘$m$’. And the force is measured in Newton ‘$N$’
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