Four equal electric charges are arranged at the four corners of a square of side 'a'. Out of these four charges, two charges are positive and placed at the ends of one diagonal. The other two charges are positive and placed at the ends of the other diagonal. The energy of the system in

Options

A. negative
B. positive
C. less than `(1)/(4piepsilon_(0))((2q^(2))/(a))`
D. greater than `(1)/(4piepsilon_(0))((2q^(2))/(a))`

Answer

To find the total energy of the system of four equal positive charges arranged at the corners of a square, we can follow these steps:

### Step 1: Understand the Configuration
We have four equal positive charges \( Q \) placed at the corners of a square with side length \( a \). The charges are arranged such that two charges are at the ends of one diagonal and the other two charges are at the ends of the other diagonal.

### Step 2: Identify the Pairs of Charges
We need to calculate the potential energy for each unique pair of charges. The pairs are:
1. Charges at corners A and B
2. Charges at corners B and C
3. Charges at corners C and D
4. Charges at corners D and A
5. Charges at corners A and C (diagonal)
6. Charges at corners B and D (diagonal)

### Step 3: Calculate the Potential Energy for Each Pair
The potential energy \( U \) between two point charges \( Q_1 \) and \( Q_2 \) separated by a distance \( r \) is given by:
\[ U = k \frac{Q_1 Q_2}{r} \]
where \( k = \frac{1}{4\pi \epsilon_0} \).

#### For adjacent charges (A-B, B-C, C-D, D-A):
- The distance \( r = a \).
- The potential energy for each pair is:
\[ U_{AB} = U_{BC} = U_{CD} = U_{DA} = k \frac{Q^2}{a} \]

#### For diagonal charges (A-C, B-D):
- The distance \( r = \sqrt{2}a \).
- The potential energy for each diagonal pair is:
\[ U_{AC} = U_{BD} = k \frac{Q^2}{\sqrt{2}a} \]

### Step 4: Total Potential Energy of the System
Now we can sum the potential energies of all pairs:
\[ U_{\text{total}} = U_{AB} + U_{BC} + U_{CD} + U_{DA} + U_{AC} + U_{BD} \]
Substituting the values:
\[ U_{\text{total}} = 4 \left( k \frac{Q^2}{a} \right) + 2 \left( k \frac{Q^2}{\sqrt{2}a} \right) \]

### Step 5: Simplifying the Expression
Factor out \( kQ^2/a \):
\[ U_{\text{total}} = k \frac{Q^2}{a} \left( 4 + 2\frac{1}{\sqrt{2}} \right) \]
Calculating \( 2/\sqrt{2} = \sqrt{2} \):
\[ U_{\text{total}} = k \frac{Q^2}{a} \left( 4 + \sqrt{2} \right) \]

### Step 6: Final Expression
Thus, the total energy of the system is:
\[ U_{\text{total}} = k \frac{Q^2}{a} (4 + \sqrt{2}) \]

### Step 7: Conclusion
Since \( k \) and \( Q^2/a \) are both positive constants, the total energy \( U_{\text{total}} \) is positive.