Two identical metal spheres A and B have equal and similar charges. They repel each other with a force 103N, when they are placed 10cm apart in a medium of dielectric constant 7. Determine the charge on each sphere.

Explanation

To determine the charge on each identical metal sphere A and B, we can use Coulomb's law, which states that the force \( F \) between two point charges is given by:

\[ F = \frac{1}{4 \pi \epsilon_0} \frac{k \cdot q^2}{r^2} \]

where:
- \( F \) is the force between the charges,
- \( k \) is the dielectric constant of the medium,
- \( q \) is the charge on each sphere,
- \( r \) is the distance between the centers of the two charges,
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Force \( F = 103 \, \text{N} \)
- Distance \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Dielectric constant \( k = 7 \)

2. **Write the Formula for the Force:**
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{k \cdot q^2}{r^2} \]

3. **Substitute the Known Values:**
\[ 103 = \frac{1}{4 \pi (8.85 \times 10^{-12})} \frac{7 \cdot q^2}{(0.1)^2} \]

4. **Calculate the Denominator:**
\[ 0.1^2 = 0.01 \]
Thus, the equation becomes:
\[ 103 = \frac{7 \cdot q^2}{4 \pi (8.85 \times 10^{-12}) \cdot 0.01} \]

5. **Calculate \( 4 \pi \epsilon_0 \cdot 0.01 \):**
\[ 4 \pi (8.85 \times 10^{-12}) \cdot 0.01 = 1.11 \times 10^{-13} \, \text{N m}^2/\text{C}^2 \]

6. **Rearranging the Equation:**
\[ 103 = \frac{7 \cdot q^2}{1.11 \times 10^{-13}} \]
\[ 103 \cdot 1.11 \times 10^{-13} = 7 \cdot q^2 \]
\[ q^2 = \frac{103 \cdot 1.11 \times 10^{-13}}{7} \]

7. **Calculate \( q^2 \):**
\[ q^2 = \frac{1.1433 \times 10^{-11}}{7} = 1.63 \times 10^{-12} \]

8. **Taking the Square Root to Find \( q \):**
\[ q = \sqrt{1.63 \times 10^{-12}} \approx 1.28 \times 10^{-6} \, \text{C} \]

9. **Final Result:**
The charge on each sphere \( q \approx 28.3 \, \mu\text{C} \) or \( 28.3 \times 10^{-6} \, \text{C} \).