Question

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:

• A. \( \frac{Kq_0^2}{a} (4\sqrt{2} - 2) \)
• B. \( \frac{Kq_0^2}{a} (4 - \sqrt{2}) \)
• C. \( \frac{Kq_0^2}{a} (3\sqrt{2} - 2) \)
• D. \( \frac{Kq_0^2}{a} (3 - \sqrt{2}) \)

Hint:

To solve problems involving potential energy, use the formula for the potential energy between two charges and sum over all pairs of charges. Pay attention to the distances between the charges in different configurations.

Answer 1

The Correct Option is A

The potential energy for a configuration of point charges is given by:

U = ∑_i<j \(\frac{K q_i q_j}{r_{ij}}\),

The distance between charges \( i \) and \( j \) is \( r_{ij} \).

Configuration (1): The charges are placed at the corners of the square. The distance between each pair of adjacent charges is \( a \), and the distance between diagonally opposite charges is \( \sqrt{2}a \).

Configuration (2): The charges are placed at the midpoints of the sides, so the distance between adjacent charges is \( \frac{a}{\sqrt{2}} \), and the distance between diagonally opposite charges is \( a \).

The potential energy of configuration (1) is:

\( U_1 = 4 \times \frac{K q_0^2}{a} + 2 \times \frac{K q_0^2}{\sqrt{2}a} \).

The potential energy of configuration (2) is:

\( U_2 = 4 \times \frac{K q_0^2}{\frac{a}{\sqrt{2}}} + 2 \times \frac{K q_0^2}{a} \).

The difference between the potential energies of configuration (2) and (1) gives the desired result:

\( \Delta U = U_2 - U_1 = \frac{K q_0^2}{a} (4\sqrt{2} - 2) \).

Final Answer: \( \frac{K q_0^2}{a} (4\sqrt{2} - 2) \).

Answer 2

Step 1: Understand the problem setup.
In configuration (1), the four charges \( q_0 \) are placed at the corners of a square with side length \( a \).
In configuration (2), the same charges are moved to the midpoints of the sides of the square.
We are asked to find the difference between the potential energies of configuration (2) and configuration (1), given that \( K = \frac{1}{4\pi \epsilon_0} \).

Step 2: Potential energy in configuration (1)
In configuration (1), the charges are placed at the corners of the square. The potential energy between two charges \( q_0 \) separated by a distance \( r \) is given by:
\[ U = \frac{K q_0^2}{r}. \] In this configuration, the charges form a square with side length \( a \), so the distances between adjacent charges are \( a \), and the diagonals of the square have a length of \( \sqrt{2}a \).
- There are 4 pairs of adjacent charges, each separated by a distance \( a \).
- There are 2 pairs of charges on the diagonals, each separated by a distance \( \sqrt{2}a \).
Thus, the potential energy in configuration (1) is:
\[ U_1 = 4 \times \frac{K q_0^2}{a} + 2 \times \frac{K q_0^2}{\sqrt{2}a} = \frac{4K q_0^2}{a} + \frac{2K q_0^2}{\sqrt{2}a}. \]

Step 3: Potential energy in configuration (2)
In configuration (2), the charges are placed at the midpoints of the sides of the square. The distance between two adjacent charges is now \( \frac{a}{\sqrt{2}} \), and the distance between the charges on the diagonals is \( a \).
- There are 4 pairs of adjacent charges, each separated by \( \frac{a}{\sqrt{2}} \).
- There are 2 pairs of charges on the diagonals, each separated by a distance \( a \).
Thus, the potential energy in configuration (2) is:
\[ U_2 = 4 \times \frac{K q_0^2}{\frac{a}{\sqrt{2}}} + 2 \times \frac{K q_0^2}{a} = \frac{4K q_0^2}{\frac{a}{\sqrt{2}}} + \frac{2K q_0^2}{a}. \] Simplifying the first term:
\[ U_2 = 4 \times \frac{K q_0^2 \sqrt{2}}{a} + \frac{2K q_0^2}{a}. \]

Step 4: Difference between potential energies
Now, we calculate the difference between the potential energies of configuration (2) and configuration (1):
\[ \Delta U = U_2 - U_1 = \left( 4 \times \frac{K q_0^2 \sqrt{2}}{a} + \frac{2K q_0^2}{a} \right) - \left( \frac{4K q_0^2}{a} + \frac{2K q_0^2}{\sqrt{2}a} \right). \] Simplifying this:
\[ \Delta U = \frac{K q_0^2}{a} \left( 4\sqrt{2} - 4 - 2 + \sqrt{2} \right). \] Simplifying further:
\[ \Delta U = \frac{K q_0^2}{a} \left( 4\sqrt{2} - 2 \right). \]

Step 5: Final answer
Thus, the difference between the potential energies of configuration (2) and configuration (1) is:
\[ \boxed{\frac{K q_0^2}{a} (4\sqrt{2} - 2)}. \]