A paisa coin is made up of $\mathrm{Al - Mg}$ alloy and weighs $0.75\, g$. It has a square shape and its diagonal measures $17$ $\mathrm{mm}$. It is electrically neutral and contains equal amounts of positive and negative charges.
A paisa coin is made up of $\mathrm{Al - Mg}$ alloy and weighs $0.75\, g$. It has a square shape and its diagonal measures $17$ $\mathrm{mm}$. It is electrically neutral and contains equal amounts of positive and negative charges.
Mass of coin $\mathrm{W}=0.79 \mathrm{~g}$
Molar mass of $\mathrm{Al}=26.9815 \mathrm{~g}$
Avogadro number $\mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23}$
No. of atoms in $26.9815 \mathrm{~g}=6.023 \times 10^{23}$
No. of atoms in $0.75 \mathrm{~g}=\mathrm{N}=?$
$\mathrm{N} =\frac{6.023 \times 10^{23} \times 0.75}{26.9815}$ $=0.16742 \times 10^{23}$ $=1.6742 \times 10^{22}$
Atomic no. of $\mathrm{Al}, \mathrm{Z}=13$. Hence, there are 13 protons and 13 electrons.
$\therefore$ Amount of positive and negative charge in one coin,
$\mathrm{Q}=\mathrm{NZe}$
$=1.6742 \times 10^{23} \times 13 \times 1.6 \times 10^{-19}$
$=3.48 \times 10^{4} \mathrm{C}$
$\therefore \mathrm{Q} =\pm 3.48 \times 10^{4} \mathrm{C}$
This amount is very large hence, we can conclude that there are positive and negative charges in large amount in neutral matter.
Similar Questions
How did Coulomb find the law of value of electric force between two point charges ?
How did Coulomb find the law of value of electric force between two point charges ?
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).
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Reason : Forces will be same in electric field
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Reason : Forces will be same in electric field
- [AIIMS 2015]
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