Two identical copper spheres are separated by 1m in vacuum. How many electrons would have to be removed from one sphere and added to the other so that they now attract each other with a force of 0.9N?

Options

A. `6.25 xx 10^(15)`
B. `6.25 xx 10^(15)`
C. `6.25 xx 10^(13)`
D. `0.65 xx 10^(13)`

Answer

To solve the problem, we need to find out how many electrons must be transferred between two identical copper spheres so that they attract each other with a force of 0.9 N when separated by 1 meter in vacuum.

### Step-by-Step Solution:

1. **Understand the Problem**:
We have two identical copper spheres, and we need to calculate how many electrons need to be transferred to create a force of attraction of 0.9 N between them.

2. **Use Coulomb's Law**:
The force \( F \) between two charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by Coulomb's law:
\[ F = k \frac{q_1 q_2}{r^2} \]
where \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

3. **Set Up the Equation**:
Since we are transferring \( n \) electrons, the charge of one electron is \( e = 1.6 \times 10^{-19} \, \text{C} \). If we remove \( n \) electrons from one sphere, it gains a charge of \( +ne \), and the other sphere, which gains those electrons, has a charge of \( -ne \). Thus, we have:
\[ q_1 = ne \quad \text{and} \quad q_2 = -ne \]

4. **Substitute into Coulomb's Law**:
Substitute \( q_1 \) and \( q_2 \) into the equation:
\[ F = k \frac{(ne)(-ne)}{r^2} = k \frac{-n^2 e^2}{r^2} \]
Since force is a magnitude, we can ignore the negative sign:
\[ F = k \frac{n^2 e^2}{r^2} \]

5. **Plug in the Values**:
We know \( F = 0.9 \, \text{N} \) and \( r = 1 \, \text{m} \):
\[ 0.9 = 9 \times 10^9 \frac{n^2 (1.6 \times 10^{-19})^2}{1^2} \]

6. **Solve for \( n^2 \)**:
Rearranging gives:
\[ n^2 = \frac{0.9 \times 1^2}{9 \times 10^9 \times (1.6 \times 10^{-19})^2} \]
Calculate \( (1.6 \times 10^{-19})^2 \):
\[ (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \]
Now substitute this value:
\[ n^2 = \frac{0.9}{9 \times 10^9 \times 2.56 \times 10^{-38}} \]
\[ n^2 = \frac{0.9}{2.304 \times 10^{-28}} \approx 3.91 \times 10^{27} \]

7. **Calculate \( n \)**:
Taking the square root:
\[ n \approx \sqrt{3.91 \times 10^{27}} \approx 6.25 \times 10^{13} \]

### Final Answer:
The number of electrons that need to be removed from one sphere and added to the other is approximately \( 6.25 \times 10^{13} \).