The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 \;\mathring A$ apart is,$\left(m_{e} \simeq 9 \times 10^{-31} kg , e=1.6 \times 10^{-19} C \right)$
(Take $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}$ )
The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 \;\mathring A$ apart is,$\left(m_{e} \simeq 9 \times 10^{-31} kg , e=1.6 \times 10^{-19} C \right)$
(Take $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}$ )
- [NEET 2020]
- A
$10^{25} \;m / s ^{2}$
- B
$10^{24} \;m / s ^{2}$
- C
$10^{23} \;m / s ^{2}$
- D
$10^{22} \;m / s ^{2}$
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$(i)$ 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).