To find the energy of the system of four equal electric charges arranged at the corners of a square, we can follow these steps:
### Step 1: Identify the Charges and Their Arrangement
We have four equal positive charges \( Q \) placed at the corners of a square with side length \( a \). Let’s denote the corners of the square as \( M, N, O, \) and \( P \).
### Step 2: Calculate the Potential Energy Between Adjacent Charges
The potential energy \( U \) between two point charges is given by the formula:
\[
U = k \frac{q_1 q_2}{r}
\]
where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between them.
For adjacent charges (e.g., \( M \) and \( N \)), the distance \( r = a \):
\[
U_{MN} = k \frac{Q \cdot Q}{a} = k \frac{Q^2}{a}
\]
Since there are four pairs of adjacent charges (MN, NO, OP, PM), the total potential energy due to adjacent charges is:
\[
U_{\text{adjacent}} = 4 \cdot k \frac{Q^2}{a} = \frac{4kQ^2}{a}
\]
### Step 3: Calculate the Potential Energy Between Diagonal Charges
For diagonal charges (e.g., \( M \) and \( O \)), the distance \( r = \sqrt{2}a \):
\[
U_{MO} = k \frac{Q \cdot Q}{\sqrt{2}a} = k \frac{Q^2}{\sqrt{2}a}
\]
Since there are two pairs of diagonal charges (MO, NP), the total potential energy due to diagonal charges is:
\[
U_{\text{diagonal}} = 2 \cdot k \frac{Q^2}{\sqrt{2}a} = \frac{2kQ^2}{\sqrt{2}a}
\]
### Step 4: Combine the Total Potential Energy
Now, we can combine the potential energy contributions from adjacent and diagonal charges:
\[
U_{\text{total}} = U_{\text{adjacent}} + U_{\text{diagonal}} = \frac{4kQ^2}{a} + \frac{2kQ^2}{\sqrt{2}a}
\]
To simplify, we can factor out \( \frac{kQ^2}{a} \):
\[
U_{\text{total}} = \frac{kQ^2}{a} \left( 4 + \frac{2}{\sqrt{2}} \right)
\]
Since \( \frac{2}{\sqrt{2}} = \sqrt{2} \), we have:
\[
U_{\text{total}} = \frac{kQ^2}{a} \left( 4 + \sqrt{2} \right)
\]
### Step 5: Substitute the Value of \( k \)
Coulomb's constant \( k \) can be expressed as:
\[
k = \frac{1}{4 \pi \epsilon_0}
\]
Substituting this into our equation gives:
\[
U_{\text{total}} = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{a} \left( 4 + \sqrt{2} \right)
\]
### Final Result
Thus, the total potential energy of the system is:
\[
U_{\text{total}} = \frac{Q^2}{4 \pi \epsilon_0 a} \left( 4 + \sqrt{2} \right)
\]
