Two sphereical bodies, each of mass 50 kg, are placed at a separation of 20 cm. Equal charges are placed on the bodies and it is found that the force of Coulomb repulsion equals the grativational attraction in magnitude. Find the magnitude of the charge placed on either body.

Explanation

To solve the problem, we need to find the magnitude of the charge placed on each of the two spherical bodies. We know that the gravitational attraction between the two bodies is equal to the Coulomb repulsion between them in magnitude.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Mass of each body, \( m = 50 \, \text{kg} \)
- Separation between the bodies, \( r = 20 \, \text{cm} = 0.2 \, \text{m} \)

2. **Write the Formula for Gravitational Force:**
The gravitational force \( F_g \) between two masses is given by:
\[ F_g = \frac{G m_1 m_2}{r^2} \]
where \( G = 6.67 \times 10^{-11} \, \text{m}^2/\text{kg} \cdot \text{s}^2 \).

Since both bodies have the same mass \( m \):
\[ F_g = \frac{G m^2}{r^2} \]

3. **Substitute the Values into the Gravitational Force Formula:**
\[ F_g = \frac{6.67 \times 10^{-11} \times (50)^2}{(0.2)^2} \]
\[ F_g = \frac{6.67 \times 10^{-11} \times 2500}{0.04} \]
\[ F_g = \frac{1.6675 \times 10^{-7}}{0.04} = 4.16875 \times 10^{-6} \, \text{N} \]

4. **Write the Formula for Coulomb Force:**
The Coulomb force \( F_c \) between two charges is given by:
\[ F_c = \frac{k q_1 q_2}{r^2} \]
where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) and since both charges are equal, \( q_1 = q_2 = q \):
\[ F_c = \frac{k q^2}{r^2} \]

5. **Set the Gravitational Force Equal to the Coulomb Force:**
Since the forces are equal in magnitude:
\[ F_g = F_c \]
\[ \frac{G m^2}{r^2} = \frac{k q^2}{r^2} \]

6. **Cancel \( r^2 \) from Both Sides:**
\[ G m^2 = k q^2 \]

7. **Rearrange to Solve for \( q^2 \):**
\[ q^2 = \frac{G m^2}{k} \]

8. **Substitute the Known Values:**
\[ q^2 = \frac{6.67 \times 10^{-11} \times (50)^2}{9 \times 10^9} \]
\[ q^2 = \frac{6.67 \times 10^{-11} \times 2500}{9 \times 10^9} \]
\[ q^2 = \frac{1.6675 \times 10^{-7}}{9 \times 10^9} = 1.85278 \times 10^{-17} \]

9. **Calculate \( q \):**
\[ q = \sqrt{1.85278 \times 10^{-17}} \approx 4.3 \times 10^{-9} \, \text{C} \]

### Final Answer:
The magnitude of the charge placed on either body is approximately \( q \approx 4.3 \times 10^{-9} \, \text{C} \).