Three point charge `Q,4Q` and `16Q` are placed on a straight line `9`cm long.Charges are placed in such a way that the system has minimum potential energy .Then

Options

A. 4Q and 16Q must be at the ends and Q at a distance of 3 cm from the 16Q
B. 4Q and 16Q must be at the ends and Q at a distance of 6 cm from the 16Q
C. Electric field at the position of Q is zero
D. Electric field at the position of Q is `Q/(4piepsilon_(0))`

Explanation

To solve the problem of arranging the point charges \( Q, 4Q, \) and \( 16Q \) on a straight line of length \( 9 \) cm for minimum potential energy, we can follow these steps:

### Step 1: Understanding the Arrangement
We need to place the charges \( Q, 4Q, \) and \( 16Q \) on a straight line such that the potential energy of the system is minimized. The potential energy \( U \) between two point charges is given by:
\[ U = k \frac{Q_1 Q_2}{r} \]
where \( k \) is Coulomb's constant, \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.

### Step 2: Placing the Largest Charges at the Ends
To minimize the potential energy, we should place the two larger charges, \( 4Q \) and \( 16Q \), at the ends of the line. This is because potential energy is inversely proportional to the distance between charges, and placing larger charges further apart will reduce their contribution to the overall potential energy.

### Step 3: Positioning the Charge \( Q \)
Let’s denote the positions of the charges as follows:
- Place \( 4Q \) at position \( x = 0 \) cm.
- Place \( 16Q \) at position \( x = 9 \) cm.
- Let the charge \( Q \) be placed at position \( x \) cm, where \( 0 < x < 9 \).

### Step 4: Finding the Distance
The distances from \( Q \) to the other charges are:
- Distance from \( 4Q \) to \( Q \): \( x \)
- Distance from \( 16Q \) to \( Q \): \( 9 - x \)

### Step 5: Writing the Total Potential Energy
The total potential energy \( U \) of the system can be expressed as:
\[ U = k \left( \frac{4Q \cdot Q}{x} + \frac{16Q \cdot Q}{9 - x} + \frac{4Q \cdot 16Q}{9} \right) \]
The last term is constant and does not depend on \( x \).

### Step 6: Minimizing the Potential Energy
To find the value of \( x \) that minimizes \( U \), we take the derivative of \( U \) with respect to \( x \) and set it to zero:
\[ \frac{dU}{dx} = -k \left( \frac{4Q^2}{x^2} - \frac{16Q^2}{(9 - x)^2} \right) = 0 \]
This leads to:
\[ \frac{4}{x^2} = \frac{16}{(9 - x)^2} \]

### Step 7: Solving the Equation
Cross-multiplying gives:
\[ 4(9 - x)^2 = 16x^2 \]
Expanding and simplifying:
\[ 4(81 - 18x + x^2) = 16x^2 \]
\[ 324 - 72x + 4x^2 = 16x^2 \]
\[ 0 = 12x^2 + 72x - 324 \]
Dividing through by 12:
\[ x^2 + 6x - 27 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-6 \pm \sqrt{36 + 108}}{2} = \frac{-6 \pm 12}{2} \]
This gives us:
\[ x = 3 \quad \text{(valid)} \quad \text{or} \quad x = -9 \quad \text{(not valid)} \]

### Step 8: Conclusion
Thus, the charge \( Q \) should be placed \( 3 \) cm from \( 4Q \) and \( 6 \) cm from \( 16Q \). Therefore, the correct arrangement is:
- \( 4Q \) at \( 0 \) cm
- \( Q \) at \( 3 \) cm
- \( 16Q \) at \( 9 \) cm