The excess (equal to number) number of electrons that must be placed on each of two small spheres spaced `3cm` apart with force of repulsion between the spheres of br `10^(-19)N` is

Options

A. `25`
B. `225`
C. `625`
D. `1250`

Explanation

To solve the problem, we need to determine the number of excess electrons that must be placed on each of the two small spheres so that the force of repulsion between them is \(10^{-19} \, \text{N}\) when they are \(3 \, \text{cm}\) apart.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Force of repulsion, \(F = 10^{-19} \, \text{N}\)
- Distance between the spheres, \(r = 3 \, \text{cm} = 0.03 \, \text{m}\)
- Charge of an electron, \(e = 1.6 \times 10^{-19} \, \text{C}\)

2. **Use Coulomb's Law**:
According to Coulomb's law, the force \(F\) between two charges \(q\) separated by a distance \(r\) is given by:
\[ F = k \frac{q_1 q_2}{r^2} \]
where \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\).

3. **Assume Charges on Each Sphere**:
Let the charge on each sphere be \(q\). Since both spheres have the same charge due to the excess electrons, we have:
\[ q_1 = q_2 = q \]

4. **Substituting Values into Coulomb's Law**:
The equation becomes:
\[ F = k \frac{q^2}{r^2} \]
Substituting the known values:
\[ 10^{-19} = 8.99 \times 10^9 \frac{q^2}{(0.03)^2} \]

5. **Rearranging to Solve for \(q^2\)**:
Rearranging gives:
\[ q^2 = \frac{10^{-19} \cdot (0.03)^2}{8.99 \times 10^9} \]
Calculate \( (0.03)^2 = 0.0009 \):
\[ q^2 = \frac{10^{-19} \cdot 0.0009}{8.99 \times 10^9} \]
\[ q^2 = \frac{9 \times 10^{-23}}{8.99 \times 10^9} \]
\[ q^2 \approx 1.0 \times 10^{-32} \]

6. **Calculating \(q\)**:
Taking the square root:
\[ q \approx 10^{-16} \, \text{C} \]

7. **Finding the Number of Electrons \(n\)**:
Since \(q = n \cdot e\), we can find \(n\):
\[ n = \frac{q}{e} = \frac{10^{-16}}{1.6 \times 10^{-19}} \]
\[ n \approx \frac{10^{-16}}{1.6 \times 10^{-19}} = 625 \]

### Final Answer:
The number of excess electrons that must be placed on each sphere is \(625\).