Suppose that one of Millikan’s oil drops has a charge of −4.8×10−19 C. How many excess electrons does the drop contain?

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Lynne Trussell · Accepted Answer

Given,One Milikian&#039;s oil drop has a charge =−4.8×10−19 CCharge of an electron =−1.6×10−19Number of electrons can be calculated as:Charge=no.of electrons×charge of an electronNo. of electrons=Chargecharge of an electronNo. of electrons=−4.8×10−19 C−1.6×10−19 C=3Answer:Number of excess electrons the drop contain =3

fudzisako · Accepted Answer

Answer:3 extra electronsExplanation:Let Q be the charge of the oil drop, which is Q=−4.8×10−19 C. The number of excess electrons, N, can be calculated as follows:N=QeSubstituting the given values, we get:N=−4.8×10−191.6×10−19Now, let&#039;s simplify this expression. When dividing two numbers in scientific notation, we subtract the exponents and divide the mantissas. Therefore:N=−4.81.6×10−1910−19Simplifying further, we have:N=−3×1Hence, the number of excess electrons on the oil drop is N=−3.Now, it&#039;s important to note that the negative sign indicates an excess of negative charge, which means the oil drop gained 3 extra electrons.

Jazz Frenia · Accepted Answer

Given:q=newhere:q is the charge of the oil drop,n is the number of excess electrons, ande is the charge of a single electron.Given that the charge of the oil drop is −4.8×10−19C, we can substitute this value into the equation:−4.8×10−19C=n·eWe know that the charge of a single electron, e, is 1.6×10−19C, so we can substitute this value as well:−4.8×10−19C=n·(1.6×10−19C)To solve for n, we can rearrange the equation:n=−4.8×10−19C1.6×10−19CSimplifying the expression, we have:n=−3Therefore, the oil drop contains −3 excess electrons. However, it&#039;s important to note that electrons are indivisible, and we cannot have a fractional number of electrons. The negative sign indicates an excess of electrons compared to the overall charge of the oil drop. In this case, it means that the oil drop is deficient in 3 electrons compared to being electrically neutral.

xleb123 · Accepted Answer

To solve the problem, we can use the equation for the charge of an electron:e=−1.6×10−19C where e represents the charge of a single electron.Given that the oil drop has a charge of −4.8×10−19C, we can calculate the number of excess electrons using the equation:Number&amp;nbsp;of&amp;nbsp;excess&amp;nbsp;electrons=Charge&amp;nbsp;of&amp;nbsp;the&amp;nbsp;oil&amp;nbsp;dropCharge&amp;nbsp;of&amp;nbsp;a&amp;nbsp;single&amp;nbsp;electronSubstituting the given values, we have:Number&amp;nbsp;of&amp;nbsp;excess&amp;nbsp;electrons=−4.8×10−19C−1.6×10−19CSimplifying the expression, we get:Number&amp;nbsp;of&amp;nbsp;excess&amp;nbsp;electrons=4.81.6=3Therefore, the oil drop contains 3 excess electrons.

Suppose that one of Millikan’s oil drops has a charge

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