A charged ball of mass `5.88xx10^(-4)` kg is suspended from two silk strings of equal lengths so that the strings are inclined at `60^(@)` with each other. Another ball carrying a charge which is equal in magnitude but opposite in sign of the first one is placed vertically below the first one at a distance of `4.2xx10^(-2) m`. Due to this, the tension in the strings is doubled. Determine the charge on the ball and the tension in the strings after electrostatic interaction. `(g=9.8 m//s^(2))`

Answer

To solve the problem step by step, we will analyze the forces acting on the charged ball and how they change when the second ball is introduced.

### Step 1: Analyze the initial equilibrium of the charged ball
- The charged ball of mass \( m = 5.88 \times 10^{-4} \, \text{kg} \) is suspended by two strings making an angle of \( 60^\circ \) with each other.
- The angle \( \theta \) of each string with the vertical is \( 30^\circ \) (since the strings are inclined at \( 60^\circ \)).
- The weight of the ball is given by \( mg \), where \( g = 9.8 \, \text{m/s}^2 \).

**Equation for vertical forces:**
\[ 2T \cos(30^\circ) = mg \]
Substituting \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ 2T \left(\frac{\sqrt{3}}{2}\right) = mg \]
\[ T \sqrt{3} = mg \]
\[ T = \frac{mg}{\sqrt{3}} \]

### Step 2: Calculate the initial tension \( T \)
Substituting the values:
\[ T = \frac{(5.88 \times 10^{-4} \, \text{kg})(9.8 \, \text{m/s}^2)}{\sqrt{3}} \]
Calculating:
\[ T = \frac{5.76 \times 10^{-3}}{1.732} \approx 3.32 \times 10^{-3} \, \text{N} \]

### Step 3: Analyze the situation after the second ball is introduced
When the second ball (with charge \(-q\)) is placed below the first ball, the tension in the strings doubles:
\[ T' = 2T \]
The new tension is \( 2T \).

### Step 4: Write the new equilibrium equation
The new equilibrium condition considering the electrostatic force \( F_e \) between the two charges is:
\[ 4T \cos(30^\circ) = mg + F_e \]
Substituting \( F_e = \frac{1}{4\pi \epsilon_0} \frac{q^2}{r^2} \) where \( r = 4.2 \times 10^{-2} \, \text{m} \):
\[ 4T \left(\frac{\sqrt{3}}{2}\right) = mg + \frac{1}{4\pi \epsilon_0} \frac{q^2}{(4.2 \times 10^{-2})^2} \]
This simplifies to:
\[ 2T \sqrt{3} = mg + \frac{1}{4\pi \epsilon_0} \frac{q^2}{(4.2 \times 10^{-2})^2} \]

### Step 5: Substitute \( T \) and simplify
From the previous steps, we know:
\[ 2T = 2 \left(\frac{mg}{\sqrt{3}}\right) \]
Substituting this into the equation:
\[ 2 \left(\frac{mg}{\sqrt{3}}\right) \sqrt{3} = mg + \frac{1}{4\pi \epsilon_0} \frac{q^2}{(4.2 \times 10^{-2})^2} \]
This simplifies to:
\[ 2mg = mg + \frac{1}{4\pi \epsilon_0} \frac{q^2}{(4.2 \times 10^{-2})^2} \]
\[ mg = \frac{1}{4\pi \epsilon_0} \frac{q^2}{(4.2 \times 10^{-2})^2} \]

### Step 6: Solve for \( q^2 \)
Rearranging gives:
\[ q^2 = mg \cdot (4.2 \times 10^{-2})^2 \cdot 4\pi \epsilon_0 \]
Substituting \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \):
\[ q^2 = (5.88 \times 10^{-4} \cdot 9.8) \cdot (4.2 \times 10^{-2})^2 \cdot 4\pi (8.85 \times 10^{-12}) \]

### Step 7: Calculate \( q \)
Calculating \( q^2 \):
\[ q^2 = (5.76 \times 10^{-3}) \cdot (1.764 \times 10^{-3}) \cdot (1.112 \times 10^{-10}) \]
\[ q^2 \approx 1.1 \times 10^{-15} \]
Thus,
\[ q \approx \sqrt{1.1 \times 10^{-15}} \approx 3.3 \times 10^{-8} \, \text{C} \]

### Final Answers
- The charge on the ball is approximately \( q \approx 3.3 \times 10^{-8} \, \text{C} \).
- The new tension in the strings is \( T' = 2T \approx 6.64 \times 10^{-3} \, \text{N} \).