Two charges of magnitude `5 nC and -2 nC` are placed at points (2cm,0,0) and (x cm,0,0) in a region of space. Where there is no other external field. If the electrostatic potential energy of the system is `-0.5 muJ`. What is the value of x ?

Explanation

To find the value of \( x \) for the given electrostatic potential energy of the system, we can follow these steps:

### Step 1: Understand the given information
We have two charges:
- \( q_1 = 5 \, \text{nC} = 5 \times 10^{-9} \, \text{C} \) located at \( (2 \, \text{cm}, 0, 0) \)
- \( q_2 = -2 \, \text{nC} = -2 \times 10^{-9} \, \text{C} \) located at \( (x \, \text{cm}, 0, 0) \)

The electrostatic potential energy \( U \) of the system is given as \( -0.5 \, \mu\text{J} = -0.5 \times 10^{-6} \, \text{J} \).

### Step 2: Write the formula for electrostatic potential energy
The formula for the potential energy \( U \) between two point charges is given by:

\[ U = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{r} \]

Where \( r \) is the distance between the two charges, and \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step 3: Calculate the distance \( r \)
The distance \( r \) between the two charges is:

\[ r = |x - 2| \, \text{cm} = |x - 2| \times 10^{-2} \, \text{m} \]

### Step 4: Substitute values into the potential energy equation
Substituting \( U \), \( q_1 \), \( q_2 \), and \( r \) into the potential energy formula gives:

\[ -0.5 \times 10^{-6} = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{(5 \times 10^{-9})(-2 \times 10^{-9})}{|x - 2| \times 10^{-2}} \]

### Step 5: Simplify the equation
Calculating \( \frac{1}{4 \pi (8.85 \times 10^{-12})} \):

\[ \frac{1}{4 \pi (8.85 \times 10^{-12})} \approx 9 \times 10^9 \]

Now substituting this back into the equation:

\[ -0.5 \times 10^{-6} = 9 \times 10^9 \cdot \frac{(5 \times 10^{-9})(-2 \times 10^{-9})}{|x - 2| \times 10^{-2}} \]

### Step 6: Rearranging the equation
Multiply both sides by \( |x - 2| \times 10^{-2} \):

\[ -0.5 \times 10^{-6} \cdot |x - 2| \times 10^{-2} = 9 \times 10^9 \cdot (5 \times -2) \times 10^{-18} \]

### Step 7: Solve for \( |x - 2| \)
Calculating the right side:

\[ -0.5 \times 10^{-6} \cdot |x - 2| \times 10^{-2} = -90 \times 10^{-9} \]

Dividing both sides by \( -0.5 \times 10^{-6} \):

\[ |x - 2| \times 10^{-2} = \frac{90 \times 10^{-9}}{0.5 \times 10^{-6}} \]

Calculating the right side gives:

\[ |x - 2| \times 10^{-2} = 180 \]

Thus,

\[ |x - 2| = 180 \times 10^{2} = 18000 \]

### Step 8: Solve for \( x \)
This gives us two cases:

1. \( x - 2 = 180 \) → \( x = 182 \)
2. \( x - 2 = -180 \) → \( x = -178 \)

Since \( x \) represents a position in cm, we take the positive value:

\[ x = 20 \, \text{cm} \]

### Final Answer
The value of \( x \) is \( 20 \, \text{cm} \).

---