Questions: Two tiny, spherical water drops, with identical charges of -8.00 x 10^-16 C, have a center-to-center separation of 0.925 cm. (a) What is the magnitude of the electrostatic force acting between them? (b) How many excess electrons are on each drop, giving it its charge imbalance?

Solution Steps

To find the magnitude of the electrostatic force between the two charged water drops, we use Coulomb's Law:

\[ F = k \frac{|q_1 q_2|}{r^2} \]

where:

• \( k = 8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2 \) is Coulomb's constant,
• \( q_1 = q_2 = -8.00 \times 10^{-16} \, \text{C} \) are the charges on the drops,
• \( r = 0.925 \, \text{cm} = 0.00925 \, \text{m} \) is the separation between the centers of the drops.

Substituting the values:

\[ F = 8.9875 \times 10^9 \frac{(-8.00 \times 10^{-16})^2}{(0.00925)^2} \]

Calculating the force:

\[ F = 8.9875 \times 10^9 \times \frac{6.4 \times 10^{-31}}{8.55625 \times 10^{-5}} \]

\[ F \approx 6.722 \times 10^{-17} \, \text{N} \]

The charge of each drop is due to excess electrons. The charge of one electron is approximately \( -1.602 \times 10^{-19} \, \text{C} \). To find the number of excess electrons \( n \) on each drop, use:

\[ n = \frac{|q|}{e} \]

where:

• \( |q| = 8.00 \times 10^{-16} \, \text{C} \),
• \( e = 1.602 \times 10^{-19} \, \text{C} \).

Substituting the values:

\[ n = \frac{8.00 \times 10^{-16}}{1.602 \times 10^{-19}} \]

Calculating the number of electrons:

\[ n \approx 4995.63 \]

Since the number of electrons must be an integer, we round to the nearest whole number:

\[ n = 4996 \]

Final Answer