Four equal charges of magnitudes `q` each are placed at four corners of a square with its centre at origin and lying in y-z plane. Aa fifth charge `+Q` is moved along x-axis. The electrostatic potential energy (U) varies on x-axis as

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Answer

To solve the problem, we need to analyze the electrostatic potential energy (U) of a charge \( +Q \) placed along the x-axis due to four equal charges \( q \) located at the corners of a square in the y-z plane. Let's break down the solution step by step.

### Step 1: Understand the Configuration
We have four charges \( q \) placed at the corners of a square with side length \( a \). The center of the square is at the origin (0, 0, 0), and the square lies in the y-z plane. The coordinates of the charges can be defined as:
- Charge 1: \( \left(0, \frac{a}{2}, \frac{a}{2}\right) \)
- Charge 2: \( \left(0, \frac{a}{2}, -\frac{a}{2}\right) \)
- Charge 3: \( \left(0, -\frac{a}{2}, \frac{a}{2}\right) \)
- Charge 4: \( \left(0, -\frac{a}{2}, -\frac{a}{2}\right) \)

The charge \( +Q \) is moved along the x-axis at a distance \( x \) from the origin.

### Step 2: Calculate the Distance from Charge \( +Q \) to Each Charge \( q \)
The distance \( r \) from the charge \( +Q \) at position \( (x, 0, 0) \) to any charge \( q \) at position \( (0, y, z) \) can be calculated using the Pythagorean theorem:
\[ r = \sqrt{x^2 + \left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2} = \sqrt{x^2 + \frac{a^2}{2}} \]

### Step 3: Calculate the Total Potential Energy
The potential energy \( U \) due to one charge \( q \) at distance \( r \) is given by:
\[ U_q = k \frac{qQ}{r} \]
where \( k \) is Coulomb's constant.

Since there are four charges, the total potential energy \( U \) is:
\[ U = 4 \cdot k \frac{qQ}{\sqrt{x^2 + \frac{a^2}{2}}} \]

### Step 4: Analyze the Variation of Potential Energy
From the expression for \( U \):
\[ U = \frac{4kqQ}{\sqrt{x^2 + \frac{a^2}{2}}} \]
we can see that as \( x \) increases, the denominator \( \sqrt{x^2 + \frac{a^2}{2}} \) increases, which means that \( U \) decreases.

### Step 5: Determine the Behavior as \( x \) Approaches Infinity
As \( x \) approaches infinity:
\[ U \to 0 \]
This indicates that the potential energy approaches zero as the charge \( +Q \) moves far away from the square of charges.

### Step 6: Find the Maximum Potential Energy
To find the maximum potential energy, we can differentiate \( U \) with respect to \( x \) and set the derivative to zero:
\[ \frac{dU}{dx} = 0 \]
Solving this will show that the maximum potential energy occurs at \( x = 0 \).

### Conclusion
The potential energy \( U \) decreases as \( x \) increases, and it is maximum when \( x = 0 \). The potential energy approaches zero as \( x \) approaches infinity.