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Check that the ratio ke^2/G me.mp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify? - CBSE (Science) Class 12 - Physics

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Question

Check that the ratio ke2/G memp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Solution

The given ratio is

`(ke^2)/(Gm_cm_p)`

Where,

G = Gravitational constant

Its unit is N mkg−2.

me and mp = Masses of electron and proton.

Their unit is kg.

e = Electric charge.

Its unit is C.

k=A constant

`=1/(4piin_0)`

0 = Permittivity of free space

Its unit is N mC−2.

therefore, unit of the given ratio `(ke^2)/(Gm_cm_p)=([Nm^2C^-2][c^-2])/([Nm^2kg^-2][kg][kg])=m^0L^0T^0`

Hence, the given ratio is dimensionless.

= 1.6 × 10−19 C

G = 6.67 × 10−11 N mkg-2

me= 9.1 × 10−31 kg

mp = 1.66 × 10−27 kg

Hence, the numerical value of the given ratio is

`(ke^2)/(Gm_cm_p)=(9xx10^9xx(1.6xx10^-19)^2)/(6.67xx10^-1xx9.1xx10^-3xx1.67xx10^-22)~~2.3xx10^39`

This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.

  Is there an error in this question or solution?

APPEARS IN

 NCERT Solution for Physics Textbook for Class 12 (2018 to Current)
Chapter 1: Electric Charge and Fields
Q: 3 | Page no. 46

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Solution Check that the ratio ke^2/G me.mp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify? Concept: Coulomb’s Law.
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