Equal charges each of `20 muC` are placed at `x = 0, 2,4,8,16 cm` on X-axis. Find the force experienced by the charge at `x = 2cm`.

Explanation

To find the force experienced by the charge at \( x = 2 \, \text{cm} \) due to the other charges located at \( x = 0 \, \text{cm}, 4 \, \text{cm}, 8 \, \text{cm}, \) and \( 16 \, \text{cm} \), we can follow these steps:

### Step 1: Identify the Charges and Their Positions
We have equal charges of \( 20 \, \mu\text{C} \) placed at the following positions on the X-axis:
- Charge \( Q_1 \) at \( x = 0 \, \text{cm} \)
- Charge \( Q_2 \) at \( x = 2 \, \text{cm} \) (the charge we are analyzing)
- Charge \( Q_3 \) at \( x = 4 \, \text{cm} \)
- Charge \( Q_4 \) at \( x = 8 \, \text{cm} \)
- Charge \( Q_5 \) at \( x = 16 \, \text{cm} \)

### Step 2: Determine the Forces Acting on Charge at \( x = 2 \, \text{cm} \)
The charge at \( x = 2 \, \text{cm} \) (let's call it \( Q_2 \)) will experience forces due to:
- \( Q_1 \) at \( x = 0 \, \text{cm} \)
- \( Q_3 \) at \( x = 4 \, \text{cm} \)
- \( Q_4 \) at \( x = 8 \, \text{cm} \)
- \( Q_5 \) at \( x = 16 \, \text{cm} \)

### Step 3: Calculate the Forces from \( Q_1 \) and \( Q_3 \)
The forces from \( Q_1 \) and \( Q_3 \) will be repulsive since all charges are positive.

1. **Force due to \( Q_1 \) (at \( x = 0 \, \text{cm} \))**:
- Distance \( r_1 = 2 \, \text{cm} = 0.02 \, \text{m} \)
- Using Coulomb's Law:
\[ F_1 = k \frac{Q_1 Q_2}{r_1^2} = k \frac{(20 \times 10^{-6})(20 \times 10^{-6})}{(0.02)^2} \]

2. **Force due to \( Q_3 \) (at \( x = 4 \, \text{cm} \))**:
- Distance \( r_2 = 2 \, \text{cm} = 0.02 \, \text{m} \)
- Using Coulomb's Law:
\[ F_2 = k \frac{Q_2 Q_3}{r_2^2} = k \frac{(20 \times 10^{-6})(20 \times 10^{-6})}{(0.02)^2} \]

### Step 4: Calculate the Forces from \( Q_4 \) and \( Q_5 \)
1. **Force due to \( Q_4 \) (at \( x = 8 \, \text{cm} \))**:
- Distance \( r_3 = 6 \, \text{cm} = 0.06 \, \text{m} \)
- Using Coulomb's Law:
\[ F_3 = k \frac{Q_2 Q_4}{r_3^2} = k \frac{(20 \times 10^{-6})(20 \times 10^{-6})}{(0.06)^2} \]

2. **Force due to \( Q_5 \) (at \( x = 16 \, \text{cm} \))**:
- Distance \( r_4 = 14 \, \text{cm} = 0.14 \, \text{m} \)
- Using Coulomb's Law:
\[ F_4 = k \frac{Q_2 Q_5}{r_4^2} = k \frac{(20 \times 10^{-6})(20 \times 10^{-6})}{(0.14)^2} \]

### Step 5: Calculate the Net Force on Charge at \( x = 2 \, \text{cm} \)
The net force \( F_{\text{net}} \) on charge \( Q_2 \) is the vector sum of all the forces acting on it:
\[ F_{\text{net}} = F_1 + F_2 - F_3 - F_4 \]
Since \( F_1 \) and \( F_2 \) are in the same direction (to the right), and \( F_3 \) and \( F_4 \) are in the opposite direction (to the left).

### Step 6: Substitute Values and Calculate
Using \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \):
\[ F_1 = F_2 = k \frac{(20 \times 10^{-6})^2}{(0.02)^2} = 9 \times 10^9 \frac{(400 \times 10^{-12})}{0.0004} = 9000 \, \text{N} \]
\[ F_3 = k \frac{(20 \times 10^{-6})^2}{(0.06)^2} = 9 \times 10^9 \frac{(400 \times 10^{-12})}{0.0036} \approx 1000 \, \text{N} \]
\[ F_4 = k \frac{(20 \times 10^{-6})^2}{(0.14)^2} = 9 \times 10^9 \frac{(400 \times 10^{-12})}{0.0196} \approx 200 \, \text{N} \]

### Final Calculation
\[ F_{\text{net}} = 9000 + 9000 - 1000 - 200 = 17000 \, \text{N} \]

### Conclusion
The net force experienced by the charge at \( x = 2 \, \text{cm} \) is approximately \( 1.2 \times 10^3 \, \text{N} \).

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