Questions: Two identical tiny conducing spheres, both initially having no charge, are brought into contact and given a total charge of 5.00 × 10^4 electrons. The spheres are then pulled apart until their centers are 12.0 cm apart. Assume the total number of electrons on the spheres remained the same as they were separated. a) What is the magnitude of the force that each sphere exerts on the other? b) Is the force attractive or repulsive?

Solution Steps

First, we need to calculate the total charge given to the spheres. The charge of one electron is approximately \( -1.602 \times 10^{-19} \, \text{C} \). Given that the total charge is \(5.00 \times 10^{4}\) electrons, the total charge in coulombs is:

\[ Q_{\text{total}} = 5.00 \times 10^{4} \times (-1.602 \times 10^{-19} \, \text{C}) \]

\[ Q_{\text{total}} = -8.01 \times 10^{-15} \, \text{C} \]

Since the spheres are identical and initially uncharged, the charge will be equally distributed between them when they are brought into contact. Therefore, the charge on each sphere is:

\[ Q = \frac{Q_{\text{total}}}{2} = \frac{-8.01 \times 10^{-15} \, \text{C}}{2} = -4.005 \times 10^{-15} \, \text{C} \]

The force between two point charges is given by Coulomb's law:

\[ F = \frac{k \cdot |Q_1 \cdot Q_2|}{r^2} \]

where:

• \(k = 8.988 \times 10^9 \, \text{N m}^2/\text{C}^2\) is Coulomb's constant,
• \(Q_1 = Q_2 = -4.005 \times 10^{-15} \, \text{C}\) are the charges on the spheres,
• \(r = 0.12 \, \text{m}\) is the distance between the centers of the spheres.

Substituting these values into the formula:

\[ F = \frac{8.988 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot (4.005 \times 10^{-15} \, \text{C})^2}{(0.12 \, \text{m})^2} \]

\[ F = \frac{8.988 \times 10^9 \cdot 1.604 \times 10^{-29}}{0.0144} \]

\[ F = \frac{1.441 \times 10^{-19}}{0.0144} \]

\[ F = 1.0007 \times 10^{-17} \, \text{N} \]

Since both spheres have the same type of charge (negative), the force between them is repulsive.

Final Answer