Chapter 21, Problem 80A

(a)

To determine

The charge on the oil drop.

Answer to Problem 80A

  $q=8.0\times {10}^{-19}\text{ C}$

Explanation of Solution

Given:

The weight of an oil drop is $F=\text{4}\text{.5}\times {10}^{-15}\text{ N}$

Intensity of the electric field is $E=5.6\times {10}^3\text{ N/C}$

Figure 1

Formula used:

If $F$ is the electrostatic force on the test charge $q$ at some point, the electric field $\left(E\right)$ on that charge at the same point is given by,

  $E=\frac{F}{q}$

  $\Rightarrow q=\frac{F}{E}$ $\left(1\right)$

Calculation:

Substituting the numerical values in equation $\left(1\right)$ ,

  $q=\frac{\text{4}\text{.5}\times {10}^{-15}\text{ N}}{5.6\times {10}^3\text{ N/C}}=8.0\times {10}^{-19}\text{ C}$

Conclusion:

The charge on the oil drop is $8.0\times {10}^{-19}\text{ C}$ .

(b)

To determine

The excess electrons carried by the oil drop.

Answer to Problem 80A

  $5$ electrons

Explanation of Solution

Given:

The weight of an oil drop is $F=\text{4}\text{.5}\times {10}^{-15}\text{ N}$

Intensity of the electric field is $E=5.6\times {10}^3\text{ N/C}$

Formula used:

The excess electrons carried by the oil drop is equal to

  $q\times \frac{1}{\text{charge of one electron}}$ $\left(2\right)$

$q=8.0\times {10}^{-19}\text{ C}$ from part (a)

Charge of one electron is $1.6\times {10}^{-19}\text{ C}$

Calculation:

Substituting the numerical values in equation $\left(2\right)$ ,

The excess electrons carried by the oil drop is equal to

  $8.0\times {10}^{-19}\text{ C}\times \frac{1}{1.6\times {10}^{-19}\text{ C}}=5$ electrons

Conclusion:

The excess electrons carried by the oil drop is $5$ electrons.