A pith ball of mass `9 xx 10^(-5)` kg carries a charge of `5muC`. What must be charge in another pith ball placed directly 2 cm above the given pith ball such that they held in equilibrium?

Options

A. `3.2 xx 10^(-11)C`
B. `7.84 xx 10^(-12)C`
C. `1.2 xx 10^(-13)C`
D. `1.6 xx 10^(-19)C`

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Calculate the weight of the first pith ball
The weight \( W \) of the first pith ball can be calculated using the formula:
\[ W = m \cdot g \]
where:
- \( m = 9 \times 10^{-5} \, \text{kg} \) (mass of the pith ball)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ W = 9 \times 10^{-5} \, \text{kg} \times 9.8 \, \text{m/s}^2 = 8.82 \times 10^{-4} \, \text{N} \]

### Step 2: Set up the equation for electrostatic force
The electrostatic force \( F \) between two charges can be calculated using Coulomb's law:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 \cdot q_2}{d^2} \]
where:
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \) (permittivity of free space)
- \( q_1 = 5 \, \mu\text{C} = 5 \times 10^{-6} \, \text{C} \) (charge of the first pith ball)
- \( d = 2 \, \text{cm} = 0.02 \, \text{m} \) (distance between the two pith balls)

Substituting the values into the formula:
\[ F = \frac{9 \times 10^9 \cdot (5 \times 10^{-6}) \cdot q_2}{(0.02)^2} \]

### Step 3: Equate weight and electrostatic force
For the pith balls to be in equilibrium, the weight of the first pith ball must equal the electrostatic force:
\[ W = F \]
Substituting the expressions we have:
\[ 8.82 \times 10^{-4} = \frac{9 \times 10^9 \cdot (5 \times 10^{-6}) \cdot q_2}{(0.02)^2} \]

### Step 4: Solve for \( q_2 \)
Rearranging the equation to solve for \( q_2 \):
\[ q_2 = \frac{8.82 \times 10^{-4} \cdot (0.02)^2}{9 \times 10^9 \cdot (5 \times 10^{-6})} \]

Calculating \( (0.02)^2 \):
\[ (0.02)^2 = 0.0004 \]

Now substituting this back into the equation:
\[ q_2 = \frac{8.82 \times 10^{-4} \cdot 0.0004}{9 \times 10^9 \cdot 5 \times 10^{-6}} \]

Calculating the numerator:
\[ 8.82 \times 10^{-4} \cdot 0.0004 = 3.528 \times 10^{-7} \]

Calculating the denominator:
\[ 9 \times 10^9 \cdot 5 \times 10^{-6} = 4.5 \times 10^4 \]

Now substituting these values:
\[ q_2 = \frac{3.528 \times 10^{-7}}{4.5 \times 10^4} = 7.84 \times 10^{-12} \, \text{C} \]

### Final Answer
Thus, the charge on the second pith ball must be:
\[ q_2 = 7.84 \times 10^{-12} \, \text{C} \]