Charge `q_(1) = +6.0n C` is on y-axis at `y = + 3cm` and charge `q_(2) = -6.0 n C` is on y-axis at `y = -3 cm`. Calculate force on a test charge `q_(0) = 2 n C` placed on X-axis at `x = 4cm`.

Explanation

To solve the problem, we need to calculate the force on a test charge \( q_0 = 2 \, \text{nC} \) placed on the x-axis at \( x = 4 \, \text{cm} \) due to two other charges \( q_1 = +6.0 \, \text{nC} \) and \( q_2 = -6.0 \, \text{nC} \) located on the y-axis at \( y = +3 \, \text{cm} \) and \( y = -3 \, \text{cm} \) respectively.

### Step 1: Determine the distances from the test charge to the other charges
- The distance from \( q_0 \) to \( q_1 \) (located at \( (0, 3) \)) can be calculated using the distance formula:
\[ r_{01} = \sqrt{(4 - 0)^2 + (0 - 3)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{cm} = 0.05 \, \text{m} \]
- The distance from \( q_0 \) to \( q_2 \) (located at \( (0, -3) \)) is the same:
\[ r_{02} = \sqrt{(4 - 0)^2 + (0 + 3)^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \, \text{cm} = 0.05 \, \text{m} \]

### Step 2: Calculate the forces exerted by \( q_1 \) and \( q_2 \) on \( q_0 \)
Using Coulomb's Law:
\[ F = k \frac{|q_1 q_0|}{r^2} \]
where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

- For \( q_1 \):
\[ F_{01} = k \frac{|q_1 q_0|}{r_{01}^2} = 9 \times 10^9 \frac{(6 \times 10^{-9})(2 \times 10^{-9})}{(0.05)^2} \]
\[ = 9 \times 10^9 \frac{12 \times 10^{-18}}{0.0025} = 9 \times 10^9 \times 4.8 \times 10^{-15} = 43.2 \times 10^{-6} \, \text{N} = 4.32 \times 10^{-5} \, \text{N} \]

- For \( q_2 \):
\[ F_{02} = k \frac{|q_2 q_0|}{r_{02}^2} = 9 \times 10^9 \frac{(6 \times 10^{-9})(2 \times 10^{-9})}{(0.05)^2} \]
\[ = 9 \times 10^9 \frac{12 \times 10^{-18}}{0.0025} = 9 \times 10^9 \times 4.8 \times 10^{-15} = 43.2 \times 10^{-6} \, \text{N} = 4.32 \times 10^{-5} \, \text{N} \]

### Step 3: Determine the direction of the forces
- The force \( F_{01} \) due to \( q_1 \) (positive charge) will be directed away from \( q_1 \) towards the right (positive x-direction).
- The force \( F_{02} \) due to \( q_2 \) (negative charge) will be directed towards \( q_2 \), which is also towards the right (positive x-direction).

### Step 4: Calculate the net force on \( q_0 \)
Since both forces are in the same direction (positive x-direction), we can add their magnitudes:
\[ F_{\text{net}} = F_{01} + F_{02} = 4.32 \times 10^{-5} \, \text{N} + 4.32 \times 10^{-5} \, \text{N} = 8.64 \times 10^{-5} \, \text{N} \]

### Step 5: Final result
The net force on the test charge \( q_0 \) is:
\[ F_{\text{net}} = 8.64 \times 10^{-5} \, \text{N} \, \text{(to the right)} \]