Question

A $1.0-mm$-diameter oil droplet (density role="math" localid="1649088632675" $900kg/m^3$ ) is negatively charged with the addition of $25$ extra electrons. It is released from rest $2.0mm$ from a very wide plane of positive charge, after which it accelerates toward the plane and collides with a speed of $3.5m/s$. What is the surface charge density of the plane?

Short answer

The surface charge density of the plane is$\mathbf{}\mathbf{6}\mathbf{.}\mathbf{4}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathit{C}\mathbf{/}\mathit{m}\mathbf{²}$

Step-by-step solution

Given information and formula used

Given :

Diameter of oil droplet : $1.0-mm$

Density : $900kg/m^3$

Charge : Negatively charged with $25$extra electrons.

It is released from rest : $2.0mm$

Collides with a speed of : $3.5m/s$.

Theory used :

Conservation of Energy says :

$\frac{1}{2}m{v}^2=qEd$

Calculating the surface charge density of the plane 

Mass of the droplet is :

$$\begin{gathered} \rho \times \frac{4}{3}{\mathrm{\pi r}}^3=900\times \frac{4}{3}\mathrm{\pi }\times {(0.5\times {10}^{-6})}^3 \\ =\underline{4.7x{10}^{-16}kg} \end{gathered}$$

Mass charge on the droplet :

$$\begin{gathered} ne=25\times 1.6\times {10}^{-19} \\ =\underline{40\times {10}^{-19}C} \end{gathered}$$

From energy conservation :

$$\begin{gathered} E=\frac{1}{2}\times \frac{mv²}{qd} \\ =\frac{1}{2}\times \frac{4.7\times {10}^{-16}\times {(3.5)}^2}{40\times {10}^{-19}\times 2\times {10}^{-3}} \\ =\underline{3.6x{10}^{5}N/C} \end{gathered}$$

We know that for electric field owing to infinite plane sheet:

$$\begin{gathered} E=\frac{\eta }{2{\epsilon }_0} \\ \Rightarrow \eta =E2{\epsilon }_0 \\ =2\times 8.85\times {10}^{-12}\times 3.6\times {10}^5 \\ =\boxed{\mathbf{}\mathbf{6}\mathbf{.}\mathbf{4}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathit{C}\mathbf{/}\mathit{m}\mathbf{²}} \end{gathered}$$