Write Coulomb’s law and explain its scalar form.
Write Coulomb’s law and explain its scalar form.
The electric force between two point charges varied inversely as the square of the distance between the charges and was directly proportional to the product of the magnitude of the two charges and acted along the line joining the two charges.
Thus, if two point charges $q_{1}, q_{2}$ are separated by a distance $r$ in vacuum, the magnitude of the force (F) between them is given by,
$\mathrm{F} \propto \frac{q_{1} q_{2}}{r^{2}}$
$\therefore \mathrm{F}=k \frac{q_{1} q_{2}}{r^{2}} \quad \ldots$ $(1)$
where, $k$ is the proportionality constant also called Coulombian constant.
Experimentally it is found that $k=8.9875 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}$. In general, it is taken as $k=9 \times$ $10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}$.
$K=\frac{1}{4 \pi \epsilon_{0}}$
where, $\epsilon_{0}=$ permittivity in free space
$=8.854185 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$
$\approx 8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$
Hence, electric force is free space,
$\mathrm{F}=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$
$\ldots$ $(2)$
If the charges are at distance ' $r$ ' in any medium instead of free space, then electric force in medium,
$\mathrm{F}_{m}=\frac{1}{4 \pi \varepsilon} \frac{q_{1} q_{2}}{r^{2}}$
where, $\varepsilon=$ permittivity in medium.
Now, relative permittivity $=\epsilon_{r}=\frac{\epsilon}{\epsilon_{0}}=k=$ dielectric constant
By taking ratio of equation (3) to (2),
$\frac{\mathrm{F}_{\mathrm{m}}}{\mathrm{F}}=\frac{\epsilon_{0}}{\epsilon}=\frac{1}{k}$
$\therefore \quad \mathrm{F}_{\mathrm{m}}=\frac{\mathrm{F}}{k}$
Thus, electric force between two charges in medium is $\frac{1}{k}$ times the electric force in free space.
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There is another useful system of units, besides the $\mathrm{SI/MKS}$. A system, called the $\mathrm{CGS}$ (centimeter-gramsecond) system. In this system Coloumb’s law is given by $\vec F = \frac{{Qq}}{{{r^2}}} \cdot \hat r$ where the distance $r$ is measured in $cm\left( { = {{10}^{ - 2}}m} \right)$ , $\mathrm{F}$ in dynes $\left( { = {{10}^{ - 5}}N} \right)$ and the charges in electrostatic units $(\mathrm{es\,unit}$), where $1$ $\mathrm{esu}$ of charge $ = \frac{1}{{[3]}} \times {10^{ - 9}}C$. The number ${[3]}$ actually arises from the speed of light in vacuum which is now taken to be exactly given by $c = 2.99792458 \times {10^8}m/s$. An approximate value of $c$ then is $c = 3 \times {10^8}m/s$.
$(i)$ Show that the coloumb law in $\mathrm{CGS}$ units yields $1$ $\mathrm{esu}$ of charge = $= 1\,(dyne)$ ${1/2}\,cm$. Obtain the dimensions of units of charge in terms of mass $\mathrm{M}$, length $\mathrm{L}$ and time $\mathrm{T}$. Show that it is given in terms of fractional powers of $\mathrm{M}$ and $\mathrm{L}$ .
$(ii)$ Write $1$ $\mathrm{esu}$ of charge $=xC$, where $x$ is a dimensionless number. Show that this gives $\frac{1}{{4\pi { \in _0}}} = \frac{{{{10}^{ - 9}}}}{{{x^2}}}\frac{{N{m^2}}}{{{C^2}}}$ with $x = \frac{1}{{[3]}} \times {10^{ - 9}}$ we have, $\frac{1}{{4\pi { \in _0}}} = {[3]^2} \times {10^9}\frac{{N{m^2}}}{{{C^2}}}$ or $\frac{1}{{4\pi { \in _0}}} = {\left( {2.99792458} \right)^2} \times {10^9}\frac{{N{m^2}}}{{{C^2}}}$ (exactly).
There is another useful system of units, besides the $\mathrm{SI/MKS}$. A system, called the $\mathrm{CGS}$ (centimeter-gramsecond) system. In this system Coloumb’s law is given by $\vec F = \frac{{Qq}}{{{r^2}}} \cdot \hat r$ where the distance $r$ is measured in $cm\left( { = {{10}^{ - 2}}m} \right)$ , $\mathrm{F}$ in dynes $\left( { = {{10}^{ - 5}}N} \right)$ and the charges in electrostatic units $(\mathrm{es\,unit}$), where $1$ $\mathrm{esu}$ of charge $ = \frac{1}{{[3]}} \times {10^{ - 9}}C$. The number ${[3]}$ actually arises from the speed of light in vacuum which is now taken to be exactly given by $c = 2.99792458 \times {10^8}m/s$. An approximate value of $c$ then is $c = 3 \times {10^8}m/s$.
$(i)$ Show that the coloumb law in $\mathrm{CGS}$ units yields $1$ $\mathrm{esu}$ of charge = $= 1\,(dyne)$ ${1/2}\,cm$. Obtain the dimensions of units of charge in terms of mass $\mathrm{M}$, length $\mathrm{L}$ and time $\mathrm{T}$. Show that it is given in terms of fractional powers of $\mathrm{M}$ and $\mathrm{L}$ .
$(ii)$ Write $1$ $\mathrm{esu}$ of charge $=xC$, where $x$ is a dimensionless number. Show that this gives $\frac{1}{{4\pi { \in _0}}} = \frac{{{{10}^{ - 9}}}}{{{x^2}}}\frac{{N{m^2}}}{{{C^2}}}$ with $x = \frac{1}{{[3]}} \times {10^{ - 9}}$ we have, $\frac{1}{{4\pi { \in _0}}} = {[3]^2} \times {10^9}\frac{{N{m^2}}}{{{C^2}}}$ or $\frac{1}{{4\pi { \in _0}}} = {\left( {2.99792458} \right)^2} \times {10^9}\frac{{N{m^2}}}{{{C^2}}}$ (exactly).