Electron Transfer Problem - Interactive

Two Insulating Spheres Problem

Two insulating small spheres are rubbed against each other and placed 1 cm apart. If they attract each other with a force of 0.1 N, then the number of electrons transferred from one sphere to the other during rubbing is given by \( x \times 10^{11} \). Find \( x \).

Select the correct option:

• A) 1.0
• B) 2.0
• C) 2.1
• D) 3.0

Detailed Explanation

The problem is solved using Coulomb's Law: \[ F = k \frac{Q^2}{r^2}, \] where:

• \( F = 0.1 \, \text{N} \),
• \( r = 1 \, \text{cm} = 0.01 \, \text{m} \),
• \( k = 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \).

Solving for \( Q \), the charge on each sphere, we have: \[ Q = \sqrt{\frac{F r^2}{k}} = \sqrt{\frac{0.1 \times (0.01)^2}{9 \times 10^9}}. \]

Calculating step-by-step:

1. Compute \( r^2 \): \( (0.01)^2 = 1 \times 10^{-4} \, \text{m}^2 \).
2. Multiply by \( F \): \( 0.1 \times 1 \times 10^{-4} = 1 \times 10^{-5} \).
3. Divide by \( k \): \[ \frac{1 \times 10^{-5}}{9 \times 10^9} \approx 1.11 \times 10^{-15}. \]
4. Taking the square root: \[ Q \approx \sqrt{1.11 \times 10^{-15}} \approx 3.33 \times 10^{-8} \, \text{C}. \]

The elementary charge (charge of an electron) is \( e = 1.6 \times 10^{-19} \, \text{C} \). Hence, the number of electrons transferred is: \[ N = \frac{Q}{e} \approx \frac{3.33 \times 10^{-8}}{1.6 \times 10^{-19}} \approx 2.08 \times 10^{11}. \]

Since the problem expresses the number as \( x \times 10^{11} \), we have \( x \approx 2.08 \), which rounds to approximately 2.1.