Chapter 21, Problem 6STP

To determine

To Choose: The correct option.

Answer to Problem 6STP

Option A

Explanation of Solution

Given:

Distance between plates, $d=63\text{mm =}\text{6}\text{.3}\times {\text{10}}^{-2}\text{m}$

Potential difference between the plates, $V=0.78\text{kV}\text{=}\text{7}\text{.80}\times {10}^2\text{V}$

Weight of the oil drop, $w=2.0\times {10}^{-14}\text{N}$

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.

Calculation:

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

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

Substitute the values and solve:

$\begin{array}{c}q=-\frac{2.0\times {10}^{-14}\text{N}\times 6.3\times {10}^{-2}\text{m}}{7.8\times {10}^2\text{V}}\\ =-1.6\times {10}^{-18}\text{C}\end{array}$

Conclusion:

Thus, option A is correct.