Chapter 21.2, Problem 32PP

To determine

The net charge on the drop and the number of excess electrons carried by it.

Answer to Problem 32PP

  $3.167\times {10}^{-19}\text{C}$ and $2$ , respectively.

Explanation of Solution

Given:

The weight of the drop, $W=1.9\times {10}^{-15}\text{N}$

The electric field acting on the drop, $E=6.0\times {10}^3\text{N/C}$

Formula used:

The electric field strength is given by,

  $E=\frac{F_{\text{on }q\text{'}}}{q\text{'}}$

Where,

  $F_{\text{on }q\text{'}}=\text{Electric force on }q\text{'}$

Calculation:

It is known that the positive charge moves in the direction of Electric field while negative charge moves in opposite direction of Electric field.

Now, the gravitational force on the drop will act in vertically downward direction, magnitude of which is equal to the weight of the ball. Thus, the electrostatic force must act in the upward direction to balance it by the gravitational force.

  $\begin{array}{c}\left|F_e\right|=\left|F_g\right|\\ \left|qE\right|=W\\ q=\frac{1.9\times {10}^{-15}}{6.0\times {10}^3}\\ q=3.167\times {10}^{-19}\text{C}\end{array}$

And, it is known as charge on electron is, $q_e=1.6\times {10}^{-19}\text{C}$ , then number of excess electrons on the drop,

  $\begin{array}{c}n=\frac{q}{q_e}\\ =\frac{3.167\times {10}^{-19}\text{C}}{1.6\times {10}^{-19}\text{C}}\\ \approx 2\end{array}$

Conclusion:

Hence, the charge on the drop is $3.167\times {10}^{-19}\text{C}$ and the number of excess electrons on the drop is $2$ .