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Coulomb’s Law

Example: Millikan Oil-Drop Experiment

A drop of oil with a weight of $3.0 \cdot 10^{-14} \text{N}$ is suspended, motionless between two plates separated by 1.5 cm. The plates are charged and have a potential difference 500 V. Find change on the drop.

Draw a Picture

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Known

$\begin{equation*} \triangle \text{V} = 500\text{V} \end{equation*}$

$\begin{equation*} \text{d} = 0.015 \text{m} (1.5 \text{cm}) \end{equation*}$

$\begin{equation*} \text{W} = 3.0 \cdot 10^{-14} \text{N} \end{equation*}$

Find

$\begin{equation*} 8\text{e}^{-} = \text{? C} \end{equation*}$

Draw a Free Body Diagram of Oil Drop

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Oil Drop is stationary so the net force on the object must be zero.

$\begin{equation*} \sum \text{F}_y = 0 \end{equation*}$

Sum of the forces:

$\begin{equation*} \text{F}_\text{e} - \text{W} = 0 \end{equation*}$

$\begin{equation*} \text{F}_\text{e} = \text{W} \end{equation*}$

note that $\text{F}_\text{e} = \text{q}_{\text{e}^{-}} \cdot \text{E}$ and $\text{E} = \frac{\triangle \text{V}}{d}$

$\begin{equation*} \left( \text{q}_{\text{e}^{-}} \right) \cdot \left( \frac{\triangle \text{V}}{d} \right) = \text{W} \end{equation*}$

Substitute parts of the equation

Solve for $\text{q}_{\text{e}^{-}}$

$\begin{equation*} \text{q}_{\text{e}^{-}} = \frac{\text{W} \cdot d}{\triangle \text{V}} \end{equation*}$

Plug in numbers and solve:

$\begin{equation*} \text{q}_{\text{e}^{-}} = \frac{3.0 \cdot 10^{-14} \cdot 0.015 \text{m}}{500 \text{V}} = \boxed{9 \cdot 10^{-19} \text{C} = \text{q}_{\text{e}^{-}}} \end{equation*}$