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  • Check that the ratio ke^2/G m_em_p is dimensionless .Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Q 1.3: Check that the ratio $\frac{k e^2}{G m_e m_p}$ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Answers (1)

best_answer

Electrostatic force 

$$
F=\frac{K Q^2}{r^2}
$$


So the dimension of

$$
\left[K e^2\right]=\left[F r^2\right]
$$


The gravitational force between two bodies of mass $M$ and $m$ is

$$
F=\frac{G M m}{r^2}
$$

so dimension of

$$
\left[G m_e m_p\right]=\left[F r^2\right]
$$


There for from (1) and (2)

$$
\left[\frac{k e^2}{G m_e m_p}\right]
$$

$\qquad$
or
Here,
$\mathrm{K}=1 / 4 \pi \epsilon_{0, \text { where }} \epsilon_0$ is the permittivity of space $\left[1 / \epsilon_0\right]=[C / V \cdot m]=\left[N m^2 C^{-2}\right]$ $e=$ Electric charge $([e]=[C])$
$\mathrm{G}=\mathrm{Gravitational}$ constant. ([G]= $\left[N \mathrm{~m}^2 \mathrm{~kg}^{-2}\right]$ ),
$m_e$ and $m_p$ are mass of electron and proton ( $m_{e]}=\left[m_{p]}=[\mathrm{Kg}]\right)$

Substituting these units, we get

$$
\left[\frac{k e^2}{G m_e m_p}\right]=\left[\frac{C^2 \times N m^2 C^{-2}}{N m^2 k g^{-2} \times k g \times k g}\right]=M^0 L^0 T^0
$$


Hence, this ratio is Dimensionless.

Putting the value of the constants

$$
\begin{aligned}
& \frac{k e^2}{G m_e m_p}=\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2}{6.67 \times 10^{11} \times 9.1 \times 10^{-31} \times 1.67 \times 10^{-27}} \\
& =2.3 \times 10^{40}
\end{aligned}
$$


The given ratio is the ratio of electric force $\frac{k e^2}{R^2}$ to the gravitational force between an electron and a proton $\frac{G m_e m_p}{R^2}$ considering the distance between them is constant!

 

 

 

Posted by

HARSH KANKARIA

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