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Home » Check that the ratio is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Check that the ratio \mathrm{ke}^{2} / \mathbf{G} \mathbf{m}_{\mathrm{e}} \mathrm{m}_{\mathrm{p}} is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
CBSE | Class 12 | Electric Charges and Fields | NCERT | Physics
Check that the ratio \mathrm{ke}^{2} / \mathbf{G} \mathbf{m}_{\mathrm{e}} \mathrm{m}_{\mathrm{p}} is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Solution:

We know that the ratio to be determined is given as follows: \frac{k e^{2}}{G m_{e} m_{p}}

where G is the gravitational constant in N \mathrm{~m}^{2} \mathrm{~kg}^{-2}

m_{e} and m_{p} is the masses of electron and proton in \mathrm{kg} .

e is the electric charge (unit -\mathrm{C})

Also, \mathrm{k}=\frac{1}{4 \pi \epsilon_{0}} \quad (unit \left.-\mathrm{Nm}^{2} \mathrm{C}^{-2}\right)

Evaluating, the unit of given ratio,

\frac{k e^{2}}{G m_{e} m_{p}}=\frac{\left[N m^{2} C^{-2}\right]\left[C^{-2}\right]}{\left[N m^{2} k g^{-2}\right][k g][k g]}=M^{0} L^{0} T^{0}

So, the given ratio is dimensionless.

Looking up in the table we have,

e=1.6 \times 10^{-19} \mathrm{C}

G=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}

\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}

m_{p}=1.66 \times 10^{-27} \mathrm{~kg}

Substituting the above values in the given ratio, we get

\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^{-2 T}}=2.3 \times 10^{38}

When the distance between a proton and an electron is constant, the electric force is equal to the gravitational force.

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