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AnswersState and explain coulomb’s inverse square law.
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Hint: The fundamental unit of electric charge is called coulomb. Coulomb’s law gives the force between two charges separated by a distance. It is found that the electric charge of any system is always an integer multiple of the least amount of charge $e$, where $e$ is the charge of the proton or electron.
Complete step by step answer:
Coulomb’s law states that the force of attraction or repulsion between two point charges is proportional to the product of charges and inversely proportional to the square of the distance between them.
If ${q_1}$ and ${q_2}$ are the two-point charges separated by a distance $r$, then by Coulomb’s law
$F \propto \dfrac{{{q_1}{q_2}}}{{{r^2}}}$ or
$F = \dfrac{{k{q_1}{q_2}}}{{{r^2}}}$,
where $k$ is the constant of proportionality given by $k = \dfrac{1}{{4\pi \varepsilon }}$.
Here $\varepsilon $ is the permittivity of the medium. If the charges are kept in free space (air or vacuum), $k = \dfrac{1}{{4\pi {\varepsilon _0}}}$, where ${\varepsilon _0}$ is the permittivity of free space.
${\varepsilon _0} = 8.854 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}$
$\therefore \dfrac{1}{{4\pi {\varepsilon _0}}} = \dfrac{1}{{4 \times 3.14 \times 8.854 \times {{10}^{ - 12}}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}$
$\therefore $ In air or vacuum, the force between two charges can be written as,
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}} = 9 \times {10^9}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Additional Information:
One coulomb is defined as the quantity of charge which when placed at a distance of $1m$ in air or vacuum from an equal and similar charge experiences a repulsive force of $9 \times {10^9}N$. If charges are similar, Coulomb force $F$ is positive and it is repulsive in nature. If one charge is positive and the other is negative, Coulomb force $F$ is negative, and it is attractive in nature.
Note: The relative permittivity or (dielectric constant) ${\varepsilon _r}$ of a medium is defined as the ratio of the permittivity of the medium to the permittivity of free space. The magnitude of Coulomb force depends on the number of charges, the distance between the charges, and the nature of the media.
Complete step by step answer:
Coulomb’s law states that the force of attraction or repulsion between two point charges is proportional to the product of charges and inversely proportional to the square of the distance between them.
If ${q_1}$ and ${q_2}$ are the two-point charges separated by a distance $r$, then by Coulomb’s law
$F \propto \dfrac{{{q_1}{q_2}}}{{{r^2}}}$ or
$F = \dfrac{{k{q_1}{q_2}}}{{{r^2}}}$,
where $k$ is the constant of proportionality given by $k = \dfrac{1}{{4\pi \varepsilon }}$.
Here $\varepsilon $ is the permittivity of the medium. If the charges are kept in free space (air or vacuum), $k = \dfrac{1}{{4\pi {\varepsilon _0}}}$, where ${\varepsilon _0}$ is the permittivity of free space.
${\varepsilon _0} = 8.854 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}$
$\therefore \dfrac{1}{{4\pi {\varepsilon _0}}} = \dfrac{1}{{4 \times 3.14 \times 8.854 \times {{10}^{ - 12}}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}$
$\therefore $ In air or vacuum, the force between two charges can be written as,
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}} = 9 \times {10^9}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Additional Information:
One coulomb is defined as the quantity of charge which when placed at a distance of $1m$ in air or vacuum from an equal and similar charge experiences a repulsive force of $9 \times {10^9}N$. If charges are similar, Coulomb force $F$ is positive and it is repulsive in nature. If one charge is positive and the other is negative, Coulomb force $F$ is negative, and it is attractive in nature.
Note: The relative permittivity or (dielectric constant) ${\varepsilon _r}$ of a medium is defined as the ratio of the permittivity of the medium to the permittivity of free space. The magnitude of Coulomb force depends on the number of charges, the distance between the charges, and the nature of the media.