A little charged bead is inside the hollow frictionless sphere manufactured from the insulting material. Sphere has a daimeter of 50 cm. The mass of the bead is 90mg, its charge is `0.5muC`. What minimum charge must carry an object at the bottom of the sphere to keep hold the charged bead at the vertax of the sphere in stable equilibrium ?

Options

A. `4.9xx10^(-8)C`
B. `9.8xx10^(-8)C`
C. `19.6xx10^(-8)C`
D. `30.2xx10^(-8)C`

Answer

To solve the problem, we need to find the minimum charge \( Q \) that must be placed at the bottom of the hollow sphere to keep the charged bead at the vertex in stable equilibrium.

### Step-by-Step Solution:

1. **Understand the Forces Acting on the Bead**:
The bead at the top of the sphere experiences two main forces:
- The gravitational force \( F_g = mg \) acting downwards.
- The electrostatic force \( F_e \) acting upwards due to the charge \( Q \) at the bottom of the sphere.

2. **Equilibrium Condition**:
For the bead to be in stable equilibrium, the upward electrostatic force must balance the downward gravitational force:
\[ F_e = mg \]

3. **Calculate the Gravitational Force**:
The mass of the bead \( m \) is given as 90 mg, which is:
\[ m = 90 \times 10^{-3} \text{ g} = 90 \times 10^{-6} \text{ kg} = 9 \times 10^{-2} \text{ kg} \]
The gravitational force is:
\[ F_g = mg = (9 \times 10^{-2} \text{ kg})(9.8 \text{ m/s}^2) = 0.882 \text{ N} \]

4. **Electrostatic Force**:
The electrostatic force between two charges is given by Coulomb's law:
\[ F_e = k \frac{|Q \cdot q|}{d^2} \]
where \( k = \frac{1}{4\pi \epsilon_0} \approx 9 \times 10^9 \text{ N m}^2/\text{C}^2 \), \( q = 0.5 \mu C = 0.5 \times 10^{-6} \text{ C} \), and \( d \) is the distance between the charges.

5. **Distance Between Charges**:
The diameter of the sphere is given as 50 cm, so the radius \( r \) is:
\[ r = \frac{50}{2} \text{ cm} = 25 \text{ cm} = 0.25 \text{ m} \]
Therefore, the distance \( d \) between the bead at the top and the charge at the bottom is \( d = 0.5 \text{ m} \).

6. **Setting Up the Equation**:
Setting the gravitational force equal to the electrostatic force gives:
\[ k \frac{|Q \cdot q|}{d^2} = mg \]

7. **Substituting Values**:
Substitute the known values into the equation:
\[ 9 \times 10^9 \frac{|Q \cdot 0.5 \times 10^{-6}|}{(0.5)^2} = 0.882 \]
Simplifying gives:
\[ 9 \times 10^9 \frac{|Q \cdot 0.5 \times 10^{-6}|}{0.25} = 0.882 \]
\[ 36 \times 10^9 |Q \cdot 0.5 \times 10^{-6}| = 0.882 \]

8. **Solving for \( Q \)**:
Rearranging gives:
\[ |Q| = \frac{0.882}{36 \times 10^9 \cdot 0.5 \times 10^{-6}} = \frac{0.882}{18 \times 10^3} \approx 4.9 \times 10^{-5} \text{ C} \]

### Final Result:
The minimum charge \( Q \) that must be placed at the bottom of the sphere to keep the bead at the vertex in stable equilibrium is approximately:
\[ Q \approx 4.9 \times 10^{-5} \text{ C} \text{ or } 49 \mu C \]