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 that must be carried by an object at the bottom of the hollow sphere to keep the charged bead at the top in stable equilibrium. Here’s a step-by-step solution:

### Step 1: Understand the Forces Acting on the Bead
The bead has a weight acting downwards due to gravity, which can be expressed as:
\[ F_{\text{weight}} = m \cdot g \]
where:
- \( m = 90 \, \text{mg} = 90 \times 10^{-6} \, \text{kg} \)
- \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Calculate the Weight of the Bead
Substituting the values into the weight formula:
\[ F_{\text{weight}} = (90 \times 10^{-6} \, \text{kg}) \cdot (9.8 \, \text{m/s}^2) \]
\[ F_{\text{weight}} = 882 \times 10^{-6} \, \text{N} = 8.82 \times 10^{-4} \, \text{N} \]

### Step 3: Determine the Electrostatic Force
The electrostatic force between two charges is given by Coulomb's law:
\[ F_{\text{electrostatic}} = k \cdot \frac{|Q_1 \cdot Q_2|}{r^2} \]
where:
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) (Coulomb's constant)
- \( Q_1 \) is the charge of the object at the bottom (unknown)
- \( Q_2 = 0.5 \, \mu\text{C} = 0.5 \times 10^{-6} \, \text{C} \)
- \( r \) is the distance between the charges, which is the diameter of the sphere (0.5 m).

### Step 4: Set the Forces Equal for Equilibrium
For the bead to be in stable equilibrium, the upward electrostatic force must equal the downward weight:
\[ F_{\text{electrostatic}} = F_{\text{weight}} \]
Thus,
\[ k \cdot \frac{|Q_1 \cdot Q_2|}{(0.5)^2} = m \cdot g \]

### Step 5: Substitute Known Values and Solve for \( Q_1 \)
Substituting the known values:
\[ 9 \times 10^9 \cdot \frac{|Q_1| \cdot (0.5 \times 10^{-6})}{(0.5)^2} = 8.82 \times 10^{-4} \]
This simplifies to:
\[ 9 \times 10^9 \cdot \frac{|Q_1| \cdot 0.5 \times 10^{-6}}{0.25} = 8.82 \times 10^{-4} \]
\[ 9 \times 10^9 \cdot 2 \cdot |Q_1| \times 10^{-6} = 8.82 \times 10^{-4} \]
\[ 18 \times 10^3 \cdot |Q_1| = 8.82 \times 10^{-4} \]
\[ |Q_1| = \frac{8.82 \times 10^{-4}}{18 \times 10^3} \]
\[ |Q_1| = \frac{8.82}{18} \times 10^{-7} \]
\[ |Q_1| = 0.49 \times 10^{-7} \, \text{C} \]
\[ |Q_1| = 4.9 \times 10^{-8} \, \text{C} \]

### Final Answer
The minimum charge that must be carried by the object at the bottom of the sphere to keep the charged bead at the top in stable equilibrium is:
\[ Q_1 = 4.9 \times 10^{-8} \, \text{C} \]