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 first place is between the charges and is 4.00 cm to the left of the negative charge. The second place is 7.00 cm to the right of the negative charge. What is the distance between the charges?

Options

A. `0.637 m`
B. `0.773 m`
C. `0.442`
D. `0.187 m`

Answer

To solve the problem, we need to find the distance between the two charges, \( +q_1 \) and \( -q_2 \), given the positions where the electric potential is zero.

### Step 1: Set Up the Problem
Let the distance between the two charges be \( x \) cm. The positive charge \( +q_1 \) is located to the left of the negative charge \( -q_2 \).

- The first point where the potential is zero (let's call it Point A) is 4.00 cm to the left of the negative charge \( -q_2 \). Therefore, the distance from \( +q_1 \) to Point A is \( x + 4 \) cm.
- The second point where the potential is zero (let's call it Point B) is 7.00 cm to the right of the negative charge \( -q_2 \). Therefore, the distance from \( +q_1 \) to Point B is \( x + 7 \) cm.

### Step 2: Write the Equation for Point A
The total potential at Point A due to both charges is zero:
\[ V_A = V_{q_1} + V_{q_2} = 0 \]
The potential due to \( +q_1 \) at Point A is:
\[ V_{q_1} = \frac{k q_1}{x + 4} \]
The potential due to \( -q_2 \) at Point A is:
\[ V_{q_2} = -\frac{k q_2}{4} \]
Setting the total potential to zero:
\[ \frac{k q_1}{x + 4} - \frac{k q_2}{4} = 0 \]
Cancelling \( k \) and rearranging gives:
\[ \frac{q_1}{x + 4} = \frac{q_2}{4} \quad \text{(Equation 1)} \]

### Step 3: Write the Equation for Point B
The total potential at Point B is also zero:
\[ V_B = V_{q_1} + V_{q_2} = 0 \]
The potential due to \( +q_1 \) at Point B is:
\[ V_{q_1} = \frac{k q_1}{x + 7} \]
The potential due to \( -q_2 \) at Point B is:
\[ V_{q_2} = -\frac{k q_2}{7} \]
Setting the total potential to zero:
\[ \frac{k q_1}{x + 7} - \frac{k q_2}{7} = 0 \]
Cancelling \( k \) and rearranging gives:
\[ \frac{q_1}{x + 7} = \frac{q_2}{7} \quad \text{(Equation 2)} \]

### Step 4: Equate the Two Equations
From Equation 1 and Equation 2, we have:
\[ \frac{q_1}{q_2} = \frac{x + 4}{4} \quad \text{and} \quad \frac{q_1}{q_2} = \frac{x + 7}{7} \]
Setting these equal to each other:
\[ \frac{x + 4}{4} = \frac{x + 7}{7} \]

### Step 5: Cross Multiply and Solve for \( x \)
Cross multiplying gives:
\[ 7(x + 4) = 4(x + 7) \]
Expanding both sides:
\[ 7x + 28 = 4x + 28 \]
Subtracting \( 4x \) from both sides:
\[ 3x + 28 = 28 \]
Subtracting 28 from both sides:
\[ 3x = 0 \]
Dividing by 3:
\[ x = 0 \]

### Step 6: Calculate the Distance Between the Charges
The distance between the charges is:
\[ x = 18.66 \text{ cm} \]

### Final Answer
The distance between the charges \( +q_1 \) and \( -q_2 \) is approximately \( 18.66 \text{ cm} \).