Chapter 21, Problem 8STP

To determine

To Calculate: The charge on the oil drop and the potential difference between the plates.

Answer to Problem 8STP

$-2.88\times {10}^{-18}\text{C}$ ; $2.99\times {10}^2\text{V}$

Explanation of Solution

Given:

Distance between plates, $d=14.1\text{mm =1}\text{.41}\times {\text{10}}^{-2}\text{m}$

Weight of the oil drop, $w=6.12\times {10}^{-14}\text{N}$

Number of extra electrons, $n=18$

Formula Used:

Electric field strength can be obtained by:

$E=\frac{V}{d}$

Here, V is the electrostatic potential difference and d is the distance between plates.

The electrostatic force:

$F=qE$

Here, q is the charge and E is the strength of the electric field.

Total charge, $Q=n{q}_e$

Here, n is the number of electrons and $q_e$ is the charge on an electron.

Calculation:

The charge on an electron is $q_e=-1.6\times {10}^{-19}\text{C}$ .

Total charge on the oil drop:

$\begin{array}{c}Q=n{q}_e\\ =18\times \left(-1.6\times {10}^{-19}\text{C}\right)\\ =-2.88\times {10}^{-18}\text{C}\end{array}$

The oil drop is suspended in the electric field. So, the electrostatic force must be equal to the weight of the oil droplet.

$\begin{array}{l}QE=-w\\ \Rightarrow Q=-\frac{w}{E}\\ \because E=\frac{V}{d}\\ \Rightarrow V=-\frac{wd}{Q}\end{array}$

Substitute the values and solve:

$\begin{array}{c}V=-\frac{6.12\times {10}^{-14}\text{N}\times 1.41\times {10}^{-2}\text{m}}{-2.88\times {10}^{-18}\text{C}}\\ =2.99\times {10}^2\text{V}\end{array}$

Conclusion:

Thus, the charge on the oil drop is $-2.88\times {10}^{-18}\text{C}$ and the potential difference between the plates is $2.99\times {10}^2\text{V}$ .