A positive charge `+q_(1)` is located to the left of a negative charge `-q_(2)` On a line passing through the two charges, there are two places where the total potential is zero. The reference is assumed to be at infinity The first place is between the charges and is 4.00 cm to the left of the negative charge. The second place is 7.00cm to the right of the negative charge if `q_(2)=-12muC` , what is the value of charge `q_(1) in muC`

Options

A. `44 muC`
B. `12 muC`
C. `32 muC`
D. `64 muC`

Explanation

To solve the problem, we need to find the value of the positive charge \( q_1 \) given that the negative charge \( q_2 = -12 \, \mu C \) and the two points where the potential is zero.

### Step-by-step Solution:

1. **Identify the Positions of the Charges:**
- Let \( q_1 \) be the positive charge located to the left of the negative charge \( q_2 \).
- The distance between the negative charge \( q_2 \) and the first point (where potential is zero) is 4 cm to the left of \( q_2 \).
- The second point (where potential is zero) is 7 cm to the right of \( q_2 \).

2. **Define the Points:**
- Let the position of \( q_2 \) be at the origin (0 cm).
- The first point (C) is at \( -4 \, cm \) (4 cm to the left of \( q_2 \)).
- The second point (D) is at \( +7 \, cm \) (7 cm to the right of \( q_2 \)).
- The distance from \( q_1 \) to point C is \( d_C = 4 + d \) (where \( d \) is the distance between \( q_1 \) and \( q_2 \)).
- The distance from \( q_1 \) to point D is \( d_D = 7 - d \).

3. **Set Up the Potential Equations:**
- The electric potential \( V \) at a point due to a charge is given by:
\[ V = k \frac{q}{r} \]
- For point C:
\[ V_C = k \frac{q_1}{d_C} + k \frac{-q_2}{4} = 0 \]
This simplifies to:
\[ \frac{q_1}{d + 4} = \frac{q_2}{4} \]
- For point D:
\[ V_D = k \frac{q_1}{d_D} + k \frac{-q_2}{7} = 0 \]
This simplifies to:
\[ \frac{q_1}{7 - d} = \frac{q_2}{7} \]

4. **Substituting \( q_2 \):**
- Substitute \( q_2 = -12 \, \mu C \) into both equations:
- From point C:
\[ \frac{q_1}{d + 4} = \frac{-12}{4} \implies \frac{q_1}{d + 4} = -3 \implies q_1 = -3(d + 4) \]
- From point D:
\[ \frac{q_1}{7 - d} = \frac{-12}{7} \implies q_1 = \frac{-12(7 - d)}{7} \]

5. **Equate the Two Expressions for \( q_1 \):**
- Set the two expressions for \( q_1 \) equal to each other:
\[ -3(d + 4) = \frac{-12(7 - d)}{7} \]
- Cross-multiply to eliminate the fraction:
\[ -21(d + 4) = -12(7 - d) \]
- Simplify:
\[ 21d + 84 = 84 - 12d \]
- Combine like terms:
\[ 33d = 0 \implies d = 0 \]
- Substitute \( d = 0 \) back into either equation to find \( q_1 \):
\[ q_1 = -3(0 + 4) = -12 \, \mu C \]

6. **Final Calculation:**
- Since \( q_1 \) is a positive charge, we take the absolute value:
\[ q_1 = 12 \, \mu C \]

### Final Answer:
The value of charge \( q_1 \) is \( 12 \, \mu C \).