It was stated and identified that we are to attempt question #51 alone.
From the given information;
The total charge on an oil drop = 5.93 × 10⁻¹⁸ C
According to Oil drop experiment carried out by Robert Milliken in (1910);
In the oil drop laboratory experiment, It was possible for Him to substitute oil for water (because water quickly evaporate). Robert Milikan adjusted the electrical voltage among the two metal plates while the oil drop continues to drop until it stopped. When the drop was stopped, the gravitational downward force on the oil drop was equivalent to the electrical upward force on the charges thereby enabling Robert Millikan to be able to estimate the electrical charge drop.
Robert Milliken identified that the magnitude of a single negative electron charge is:
= 1.6 × 10⁻¹⁹ C
∴
If in an experiment, the total charge = 5.93 × 10⁻¹⁸ C
Then;
The number of negative charges in the oil drop can be calculated as;
[tex]\mathbf{=\dfrac{5.93 \times 10^{-18}}{1.6 \times 10^{-19}}}[/tex]
= 37.063 negative charges
Therefore, we can conclude that the numbers of negative charges contained in the total drop = 37.063.
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