Question

In a Millikan's oil drop experiment , an oil drop of mass $0.64\times {10}^{-14}\text{kg}$ carrying a charge $1.6\times {10}^{-19}\text{C}$ remains stationary between two plates separated by a distance of $5\text{mm}.$ The voltage that must be applied between the plates is-
[Take $g=9.8{\text{ms}}^{-2}$ ]

• A. $1960\text{V}$
• B. $980\text{V}$
• C. $3920\text{V}$
• D. $2880\text{V}$

Answer 1

The correct option is A $1960\text{V}$

Given,

$m=0.64\times {10}^{-14}\text{kg};q=1.6\times {10}^{-19}\text{C}d=5\text{mm};g=9.8{\text{ms}}^{-2}$

Since, the oil drop is at rest,


$\text{Force due to electric field = Weight of the oil drop}$

$Eq=mg$

$(\frac{V}{d})q=mg$

$V=\frac{mgd}{q}=\frac{64\times {10}^{-16}\times 9.8\times 5\times {10}^{-3}}{1.6\times {10}^{-19}}$

$V=200\times 9.8=1960\text{V}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.