91Ó°ÊÓ

1. The charge on the drop,$q=ne$
2. The given data table

Using the concept of electric charge, we can get the required values of n for each value of charge, q. Again, by creating the approximation value of the charge, we can get the required value of the charge.

Formula:

The value of charge of number of atoms, $q=ne$ (i)

If we subtract each value from the next larger value in the table, we find a set of numbers that are suggestive of a basic unit of charge:, $1.64x{10}^{−19}$,$3.3x{10}^{−19}$,$1.63x{10}^{−19}$ , $3.35x{10}^{−19}$, $1.6x{10}^{−19}$,$1.63x{10}^{−19}$ , $3.18x{10}^{−19}$,$3.24x{10}^{−19}$where the SI unit Coulomb is understood. These values are either close to a common charge value, $e=1.6x{10}^{−19}C$or are double that. Taking this, then, as a crude approximation to our experimental e, we divide it into all the values in the original data set using equation (i) and round to the nearest integer, obtaining $n=4,5,7,8,10,11,12,14,$and.$16$

When we perform a least squares fit of the original data set versus these values for n we obtain the linear equation:

$q=\text{(}7.18x{10}^{−21}+1.633x{10}^{−19}\text{n})\text{C}$

If we dismiss the constant term as unphysical (representing, say, systematic errors in our measurements) then we obtain$e=1.63x{10}^{−19}\text{ C}$when we set$n=1$in this equation.

Hence, the linear regression fits the value and the value of the charge is.$1.63x{10}^{−19}\text{ C}$