Quiz 3: Two identical charges of 2.0 µC are placed at the vertices of a square with side lengths of 2.0 m. Calculate the electric potential at the center of the square (assuming the potential is zero at infinity).
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To calculate the electric potential at the center of a square due to two identical charges placed at its vertices, you can use the formula for the electric potential due to a point charge, which is:
\[ V = k \frac{q}{r} \]
where \( V \) is the electric potential, \( k \) is Coulomb's constant (\( k = 8.988 \times 10^9 \ \text{N} \cdot \text{m}^2/\text{C}^2 \)), \( q \) is the charge, and \( r \) is the distance from the charge to the point where you're calculating the potential.
Here are the logical steps to find the electric potential at the center of the square:
Step 1: Calculate the distance from the vertices to the center of the square.
Since the charges are at opposite vertices of a square with side length of 2.0 m, the diagonal of the square will be \( \text{diagonal} = \sqrt{2} \times \text{side length} = \sqrt{2} \times 2.0 \ \text{m} \).
The center of the square is halfway along the diagonal, so the distance from a vertex to the center is half the diagonal:
\[ r = \frac{\text{diagonal}}{2} = \frac{\sqrt{2} \times 2.0 \ \text{m}}{2} = \sqrt{2} \ \text{m} \]
Step 2: Calculate the electric potential due to a single charge at the center.
For one charge of 2.0 µC, the electric potential at the center is:
\[ V_1 = k \frac{q}{r} = 8.988 \times 10^9 \ \text{N} \cdot \text{m}^2/\text{C}^2 \times \frac{2.0 \times 10^{-6} \ \text{C}}{\sqrt{2} \ \text{m}} \]
Step 3: Since we have two identical charges, the total potential at the center is simply twice the potential due to one charge because electric potential is a scalar quantity and can be added directly:
\[ V_{total} = 2V_1 \]
\[ V_{total} = 2 \times 8.988 \times 10^9 \times \frac{2.0 \times 10^{-6}}{\sqrt{2}} \]
\[ V_{total} = 2 \times 8.988 \times 10^9 \times \frac{2.0 \times 10^{-6}}{1.414} \]
\[ V_{total} = \frac{2 \times 8.988 \times 10^9 \times 2.0 \times 10^{-6}}{1.414} \]
\[ V_{total} = \frac{35.952 \times 10^3}{1.414} \ \text{V} \]
\[ V_{total} ≈ 25.43 \times 10^3 \ \text{V} \]
\[ V_{total} ≈ 25.43 \ \text{kV} \]
So, the electric potential at the center of the square is approximately 25.43 kV.