Question

Two small objects $X$ and $Y$ are permanently separated by a distance 1 cm. Object $X$ has a charge of +1.0$\mu C$ and object $Y$ has a charge of $-1.0\mu C$. A certain number of electrons are removed from $X$ and put onto $Y$ to make the electrostatic force between the two objects an attractive force whose magnitude is 360 N. Number of electrons removed is

• A. 8.4x$10^{13}$
• B. 6.25x$10^{12}$
• C. 4.2x$10^{11}$
• D. 3.5x$10^{10}$

Answer 1

Answer: (B)

6.25x$10^{12}$

Answer 2

To determine the number of electrons transferred, we apply Coulomb's Law.

1. Define initial conditions:

   • Object $X$ has a charge of $+1.0\,\mu\text{C}$.
   • Object $Y$ has a charge of $-1.0\,\mu\text{C}$.

2. Account for electron transfer:

   • Let $n$ be the number of electrons transferred from $X$ to $Y$.
   • The charge of each electron is $e = 1.6 \times 10^{-19}\,\text{C}$.
   • The new charge on $X$ is $Q_X = 1.0 \times 10^{-6} + ne$.
   • The new charge on $Y$ is $Q_Y = -1.0 \times 10^{-6} - ne$.

3. Apply Coulomb's Law:

   • $F = k \frac{|Q_X Q_Y|}{r^2}$, where $F = 360\,\text{N}$, $r = 0.01\,\text{m}$, and $k = 9.0 \times 10^9\, \text{N m}^2/\text{C}^2$.
   • $360 = \frac{9.0 \times 10^9 \left(1.0 \times 10^{-6} + ne\right)^2}{(0.01)^2}$.

4. Solve for $ne$:

   • $\left(1.0 \times 10^{-6} + ne\right)^2 = \frac{360 \times (0.01)^2}{9.0 \times 10^9} = 4.0 \times 10^{-12}$.
   • $1.0 \times 10^{-6} + ne = \sqrt{4.0 \times 10^{-12}} = 2.0 \times 10^{-6}$.
   • $ne = 2.0 \times 10^{-6} - 1.0 \times 10^{-6} = 1.0 \times 10^{-6}\,\text{C}$.

5. Solve for $n$:

   • $n = \frac{1.0 \times 10^{-6}}{1.6 \times 10^{-19}} \approx 6.25 \times 10^{12}$.

Therefore, the number of electrons removed from $X$ and put onto $Y$ is approximately $6.25 \times 10^{12}$.