# 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? - Physics

Numerical

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"_"e""m"_"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"_"e""m"_"p") = (["Nm"^2"C"^-2]["C"^-2])/(["Nm"^2"kg"^-2]["kg"]["kg"]) = "m"^0"L"^0"T"^0

Hence, the given ratio is dimensionless.

e = 1.6 × 10−19 C

G = 6.67 × 10−11 N mkg2

me = 9.1 × 10−31 kg

mp = 1.66 × 10−27 kg

Hence, the numerical value of the given ratio is

("ke"^2)/("Gm"_"e""m"_"p") =(9 xx 10^9 xx (1.6 xx 10^-19)^2)/(6.67 xx 10^-11 xx 9.1 xx 10^-31 xx 1.67 xx 10^-27) ≈ 2.3 xx 10^39

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

Concept: Coulomb’s Law - Force Between Two Point Charges
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 1 Electric Charges and Fields
Exercise | Q 1.3 | Page 46
NCERT Class 12 Physics Textbook
Chapter 1 Electric Charge and Fields
Exercise | Q 3 | Page 46
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