Three point charges q,`q//3` and 16q have to be arranged on positive x-axis within 20 cm, so that system's potential energy is minimum. Find the distance of charge `q//3` from charge q.

Answer

To solve the problem of arranging the three point charges \( q \), \( \frac{q}{3} \), and \( 16q \) on the positive x-axis such that the system's potential energy is minimized, we can follow these steps:

### Step 1: Define the Arrangement
Let's place the charges on the x-axis:
- Charge \( q \) at position \( x_1 = 0 \) cm.
- Charge \( \frac{q}{3} \) at position \( x_2 = x \) cm.
- Charge \( 16q \) at position \( x_3 = 20 \) cm.

### Step 2: Calculate the Potential Energy
The potential energy \( U \) of the system of charges can be expressed as the sum of the potential energies between each pair of charges:

\[ U = U_{12} + U_{23} + U_{31} \]

Where:
- \( U_{12} = k \frac{q \cdot \frac{q}{3}}{x} = \frac{kq^2}{3x} \)
- \( U_{23} = k \frac{\frac{q}{3} \cdot 16q}{20 - x} = \frac{16kq^2}{3(20 - x)} \)
- \( U_{31} = k \frac{q \cdot 16q}{20} = \frac{16kq^2}{20} \)

Thus, the total potential energy is:

\[ U = \frac{kq^2}{3x} + \frac{16kq^2}{3(20 - x)} + \frac{16kq^2}{20} \]

### Step 3: Simplify the Expression
We can factor out \( kq^2 \):

\[ U = kq^2 \left( \frac{1}{3x} + \frac{16}{3(20 - x)} + \frac{16}{20} \right) \]

### Step 4: Differentiate the Potential Energy
To find the minimum potential energy, we need to differentiate \( U \) with respect to \( x \) and set the derivative to zero:

\[ \frac{dU}{dx} = -\frac{kq^2}{3x^2} + \frac{16kq^2}{3(20 - x)^2} \]

Setting \( \frac{dU}{dx} = 0 \):

\[ -\frac{1}{3x^2} + \frac{16}{3(20 - x)^2} = 0 \]

### Step 5: Solve for \( x \)
Cross-multiplying gives:

\[ (20 - x)^2 = 16x^2 \]

Taking the square root:

\[ 20 - x = 4x \quad \text{or} \quad 20 - x = -4x \]

From \( 20 - x = 4x \):

\[ 20 = 5x \quad \Rightarrow \quad x = 4 \text{ cm} \]

From \( 20 - x = -4x \):

\[ 20 = -3x \quad \Rightarrow \quad x = -\frac{20}{3} \text{ (not valid since distance cannot be negative)} \]

### Step 6: Verify Minimum
To ensure that this value of \( x \) gives a minimum potential energy, we can check the second derivative:

\[ \frac{d^2U}{dx^2} > 0 \text{ at } x = 4 \]

This confirms that \( x = 4 \) cm is indeed the point where potential energy is minimized.

### Final Answer
The distance of charge \( \frac{q}{3} \) from charge \( q \) is:

\[ \boxed{4 \text{ cm}} \]