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Question: Prove 1 Debye = \({10^{ - 18}}esu\) cm =\(3.335 \times {10^{30}}C - m\)( coulomb meter)....

Prove 1 Debye = 10−18esu{10^{ - 18}}esu cm =3.335×1030C−m3.335 \times {10^{30}}C - m( coulomb meter).

Explanation

Solution

In order to solve the question, we will first use the formula of dipole moment then we will substitute the value of distance and charge used to calculate the 1 Debye after then we will convert the units into the units which are asked in the question and then we will reach at the answer.

Formula used:
μ=q×d\mu = q \times d
μ\mu = Dipole moment
q = charge
d = distance

Complete step by step answer:
In the question are asked to prove that 1 Debye = 10−18esu{10^{ - 18}}esu cm =3.335×1030C−m3.335 \times {10^{30}}C - m( coulomb meter)
A dipole is pair of two opposite and equal charges separated by a small distance and both the charges remains together because of the attractive forces
Dipole moment (μ\mu ) is created by the dipoles they are the product of charge and the distance by which they both are separated
μ=q×d\mu = q \times d
Historically Debye was defined as the dipole moment which is the result of two charges of opposite sign but an equal magnitude
The charge was considered as 10−10{10^{ - 10}} star coulomb which was used in the calculation of dipole moment of 1 Debye and the specific distance by which charge were separated that was 1 angstrom
1 angstrom = 10−8{10^{ - 8}}cm
Star coulomb was generally called e.s.u. and e.s.u. stands for electrostatic unit e.s.u. was generally used in the older literature
For calculating the 1 Debye we will use the formula
μ=q×d\mu = q \times d
Substituting the value of d and q
1 Debye = 10−18esu cm1{\text{ Debye = }}{10^{ - 18}}esu{\text{ cm}}
Now we change the units in coulomb and m
1 esu = 3.335×10−10C1{\text{ }}esu{\text{ = 3}}{\text{.335}} \times {10^{ - 10}}C
1cm = 10−2m1cm{\text{ = 1}}{{\text{0}}^{ - 2}}m
Substituting the values in the 1 Debye
1 Debye = 10−18×3.335×10−10 C×10−2m1{\text{ Debye = }}{10^{ - 18}} \times {\text{3}}{\text{.335}} \times {10^{ - 10}}{\text{ C}} \times {10^{ - 2}}m
1 Debye = 3.335×10−30 C - m1{\text{ Debye = 3}}{\text{.335}} \times {10^{ - 30}}{\text{ C - }}m
Therefore, 1 Debye = 10−18esu cm = 3.335×10−30 C - m1{\text{ Debye = }}{10^{ - 18}}esu{\text{ cm = 3}}{\text{.335}} \times {10^{ - 30}}{\text{ C - }}m.

Note: Many of the people may be confused as star coulomb has a less mention in the theory as it was used in the ancient times along with this e.s.u. was also used a long time back therefore as they both were used as measure of charge that is why they have relation with coulomb which is used nowadays as measure of charge.