Question

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

Answer 1

Hint: 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:
 $\mu = q \times d$
$\mu $ = Dipole moment
q = charge
d = distance

Complete step by step answer:
In the question are asked to prove that 1 Debye = ${10^{ - 18}}esu$ cm =$3.335 \times {10^{30}}C - m$( 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 ($\mu $) is created by the dipoles they are the product of charge and the distance by which they both are separated
$\mu = q \times d$
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 ${10^{ - 10}}$ 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 = ${10^{ - 8}}$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
$\mu = q \times d$
Substituting the value of d and q
$1{\text{ Debye = }}{10^{ - 18}}esu{\text{ cm}}$
Now we change the units in coulomb and m
$1{\text{ }}esu{\text{ = 3}}{\text{.335}} \times {10^{ - 10}}C$
$1cm{\text{ = 1}}{{\text{0}}^{ - 2}}m$
Substituting the values in the 1 Debye
$1{\text{ Debye = }}{10^{ - 18}} \times {\text{3}}{\text{.335}} \times {10^{ - 10}}{\text{ C}} \times {10^{ - 2}}m$
$1{\text{ Debye = 3}}{\text{.335}} \times {10^{ - 30}}{\text{ C - }}m$
Therefore, $1{\text{ Debye = }}{10^{ - 18}}esu{\text{ cm = 3}}{\text{.335}} \times {10^{ - 30}}{\text{ C - }}m$.

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.