Question

Two small insulating spheres are rubbed against each other and placed \(1\,\text{cm}\) apart. If they attract each other with a force \(F=0.1\,\text{N}\), the number of electrons that were transferred from one sphere to another during rubbing are nearly

• A. \(5\times10^{11}\)
• B. \(2\times10^{11}\)
• C. \(2\times10^{10}\)
• D. \(1\times10^{12}\)

Hint:

For identical charges produced by rubbing: \[ F=\frac{kq^2}{r^2} \] First find \(q\), then use \[ n=\frac{q}{e} \] where \[ e=1.6\times10^{-19}\,\text{C} \] to determine the number of transferred electrons.

Answer 1

The Correct Option is B

Concept: When electrons are transferred from one insulating sphere to another, the spheres acquire equal and opposite charges. According to Coulomb's law, \[ F=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{r^2} \] where \[ \frac{1}{4\pi\varepsilon_0}=9\times10^9\ \text{N m}^2\text{C}^{-2} \]

Step 1: Calculate the charge on each sphere. Given, \[ F=0.1\,\text{N} \] \[ r=1\,\text{cm}=10^{-2}\,\text{m} \] Using Coulomb's law, \[ 0.1 = 9\times10^9 \frac{q^2}{(10^{-2})^2} \] \[ 0.1 = 9\times10^{13}q^2 \] \[ q^2 = \frac{0.1}{9\times10^{13}} \] \[ q^2 = 1.11\times10^{-15} \] \[ q = 3.33\times10^{-8}\,\text{C} \]

Step 2: Find the number of electrons transferred. \[ q=ne \] where \[ e=1.6\times10^{-19}\,\text{C} \] Therefore, \[ n=\frac{q}{e} \] \[ n= \frac{3.33\times10^{-8}} {1.6\times10^{-19}} \] \[ n = 2.08\times10^{11} \] \[ n\approx2\times10^{11} \]

Step 3: State the answer. \[ \boxed{ n = 2\times10^{11}\ \text{electrons} } \] Hence, the correct option is \[ \boxed{(B)} \]