Two point charges are fixed on the Xaxis , `q_(1)=12.0mC` is located at the origin and `q_(2)=-3.0mC` is located at point A, with `x_(2)=8.0cm`. Where should a third charge `q_(3)`, be placed on the X-axis so that the total, electrostatic force acting on it is zero?

Options

A. at `x_(3)=16cm`
B. at `x_(3)=12cm`
C. at `x_(3)=24cm`
D. at `x_(3)=6cm`

Explanation

To solve the problem of where to place the third charge \( q_3 \) so that the total electrostatic force acting on it is zero, we can follow these steps:

### Step 1: Understand the Configuration
We have two charges:
- \( q_1 = 12.0 \, \text{mC} \) (positive) located at the origin \( (0, 0) \).
- \( q_2 = -3.0 \, \text{mC} \) (negative) located at \( (8.0 \, \text{cm}, 0) \).

We need to find the position of a third charge \( q_3 \) on the x-axis such that the net electrostatic force acting on it is zero.

### Step 2: Analyze Possible Positions for \( q_3 \)
The third charge \( q_3 \) can be placed in three possible regions:
1. To the left of \( q_1 \) (i.e., \( x < 0 \))
2. Between \( q_1 \) and \( q_2 \) (i.e., \( 0 < x < 8 \))
3. To the right of \( q_2 \) (i.e., \( x > 8 \))

### Step 3: Evaluate Each Region
1. **Left of \( q_1 \)**:
- If \( q_3 \) is positive, it will be repelled by \( q_1 \) and attracted by \( q_2 \). The forces will act in opposite directions, potentially balancing each other.

2. **Between \( q_1 \) and \( q_2 \)**:
- If \( q_3 \) is positive, it will be repelled by \( q_1 \) and attracted by \( q_2 \). The net force cannot be zero since both forces will act in the same direction.

3. **Right of \( q_2 \)**:
- If \( q_3 \) is positive, it will be repelled by \( q_2 \) and attracted by \( q_1 \). The forces will act in opposite directions, potentially balancing each other.

### Step 4: Set Up the Equation for Forces
Assuming \( q_3 \) is placed at a distance \( x \) from the origin (to the left of \( q_1 \)), the distances to the charges are:
- Distance from \( q_1 \): \( |x| \)
- Distance from \( q_2 \): \( 8 + |x| \)

Using Coulomb's Law, we can express the forces:
- Force due to \( q_1 \) on \( q_3 \):
\[ F_1 = k \frac{|q_1 \cdot q_3|}{x^2} \]
- Force due to \( q_2 \) on \( q_3 \):
\[ F_2 = k \frac{|q_2 \cdot q_3|}{(8 + |x|)^2} \]

Setting the magnitudes of the forces equal for equilibrium:
\[ k \frac{12 \cdot q_3}{x^2} = k \frac{3 \cdot q_3}{(8 + |x|)^2} \]

### Step 5: Simplify the Equation
Since \( k \) and \( q_3 \) are common on both sides, they can be canceled out:
\[ \frac{12}{x^2} = \frac{3}{(8 + |x|)^2} \]

Cross-multiplying gives:
\[ 12(8 + |x|)^2 = 3x^2 \]

### Step 6: Expand and Rearrange
Expanding the left side:
\[ 12(64 + 16|x| + x^2) = 3x^2 \]
\[ 768 + 192|x| + 12x^2 = 3x^2 \]
Rearranging gives:
\[ 9x^2 + 192|x| + 768 = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 9, b = 192, c = 768 \):
\[ x = \frac{-192 \pm \sqrt{192^2 - 4 \cdot 9 \cdot 768}}{2 \cdot 9} \]

Calculating the discriminant:
\[ 192^2 = 36864 \]
\[ 4 \cdot 9 \cdot 768 = 27648 \]
\[ \sqrt{36864 - 27648} = \sqrt{9216} = 96 \]

Now substituting back:
\[ x = \frac{-192 \pm 96}{18} \]
Calculating the two possible values:
1. \( x = \frac{-96}{18} = -5.33 \, \text{cm} \)
2. \( x = \frac{-288}{18} = -16 \, \text{cm} \)

### Conclusion
The valid position for \( q_3 \) is \( x = -16 \, \text{cm} \) (to the left of the origin).