Two pith balls,, each weighting 10mg are suspended from the sam point by silk threads, each of length 0.25m. When equal and similar charges are placed on them they repel each other and are 10m apart. Find the charge on the each pith ball.

Answer

To solve the problem step by step, we will analyze the forces acting on the pith balls and apply the principles of electrostatics and trigonometry.

### Step 1: Understand the setup
We have two pith balls, each weighing 10 mg (which is \(10 \times 10^{-3}\) g = \(10 \times 10^{-6}\) kg). They are suspended from the same point by silk threads of length 0.25 m. When charged, they repel each other and are 10 cm apart.

### Step 2: Convert units
Convert the mass of the pith balls to kilograms:
\[ m = 10 \, \text{mg} = 10 \times 10^{-6} \, \text{kg} = 10^{-5} \, \text{kg} \]

### Step 3: Identify forces acting on the pith balls
1. The weight (\(W\)) acting downwards:
\[ W = mg = 10^{-5} \times 9.81 \approx 9.81 \times 10^{-5} \, \text{N} \]
2. The tension (\(T\)) in the thread acting upwards.
3. The electrostatic force (\(F_e\)) acting horizontally due to the charges.

### Step 4: Geometry of the setup
When the pith balls are charged, they form an isosceles triangle with the threads. The distance between the two balls is 10 cm (0.1 m), and each thread is 0.25 m long.

### Step 5: Calculate the angle
Let \(d\) be the distance between the two pith balls (0.1 m), and \(L\) be the length of the thread (0.25 m). The horizontal distance from the suspension point to each ball is:
\[ x = \frac{d}{2} = \frac{0.1}{2} = 0.05 \, \text{m} \]

Using the Pythagorean theorem:
\[ L^2 = x^2 + h^2 \quad \text{(where \(h\) is the vertical distance)} \]
\[ 0.25^2 = 0.05^2 + h^2 \]
\[ 0.0625 = 0.0025 + h^2 \]
\[ h^2 = 0.0625 - 0.0025 = 0.06 \implies h = \sqrt{0.06} \approx 0.245 \, \text{m} \]

### Step 6: Resolve forces
1. The vertical component of tension balances the weight:
\[ T \cos \theta = mg \]
2. The horizontal component of tension equals the electrostatic force:
\[ T \sin \theta = F_e \]

### Step 7: Calculate \(\tan \theta\)
From the triangle:
\[ \tan \theta = \frac{x}{h} = \frac{0.05}{0.245} \approx 0.204 \]

### Step 8: Relate the forces
Using the tangent:
\[ \frac{F_e}{mg} = \tan \theta \]
Thus,
\[ F_e = mg \tan \theta \]

### Step 9: Calculate \(F_e\)
Substituting the values:
\[ F_e = 9.81 \times 10^{-5} \times 0.204 \approx 2.00 \times 10^{-5} \, \text{N} \]

### Step 10: Use Coulomb's Law
According to Coulomb's law:
\[ F_e = k \frac{q^2}{d^2} \]
Where \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\) and \(d = 0.1 \, \text{m}\):
\[ 2.00 \times 10^{-5} = 9 \times 10^9 \frac{q^2}{(0.1)^2} \]
\[ 2.00 \times 10^{-5} = 9 \times 10^9 \frac{q^2}{0.01} \]
\[ 2.00 \times 10^{-5} = 9 \times 10^{11} q^2 \]
\[ q^2 = \frac{2.00 \times 10^{-5}}{9 \times 10^{11}} \approx 2.22 \times 10^{-17} \]
\[ q \approx \sqrt{2.22 \times 10^{-17}} \approx 4.7 \times 10^{-9} \, \text{C} \]

### Final Answer
The charge on each pith ball is approximately \(4.7 \times 10^{-9} \, \text{C}\).