To compare the electric potentials at the four points (A, B, C, and D) due to the charges at the corners of the square, we need to consider the contributions from each charge to the potential at each point.
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Electric potential (V) due to a point charge is given by the formula: \[ V = k \frac{Q}{r} \] where \(k\) is Coulomb's constant, \(Q\) is the charge, and \(r\) is the distance from the charge to the point where the potential is being calculated.
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Charge contributions:
- Charge \(Q\) is located at point A (bottom left corner).
- Charge \(2Q\) is located at point B (bottom right corner).
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Calculating potential at each point:
- Point A (top left corner):
- Peas from Q: \[ V_A = k \frac{Q}{d} + k \frac{2Q}{\sqrt{2}d} \quad (\text{distance to 2Q is } \sqrt{(d^2 + d^2)} = \sqrt{2}d) \]
- Point B (top right corner):
- Contribution from Q: \[ V_B = k \frac{Q}{d} + k \frac{2Q}{d} \]
- Point C (center of the square):
- Contributions from both charges: \[ V_C = k \frac{Q}{\frac{d}{\sqrt{2}}} + k \frac{2Q}{\frac{d}{\sqrt{2}}} \] \[ V_C = k \frac{3Q}{\frac{d}{\sqrt{2}}} = k \frac{3\sqrt{2}Q}{d} \]
- Point D (center of bottom edge):
- Contributions from both charges: \[ V_D = k \frac{Q}{\frac{d}{2}} + k \frac{2Q}{\frac{d}{2}} = k \frac{3Q}{\frac{d}{2}} = k \frac{6Q}{d} \]
- Point A (top left corner):
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Comparing potentials:
- \( V_A \) has contributions from \( k \frac{Q}{d} + k \frac{Q\sqrt{2}}{d} \)
- \( V_B = k \frac{3Q}{d} \)
- \( V_C \) is \( k \frac{3\sqrt{2}Q}{d} \)
- \( V_D = k \frac{6Q}{d} \)
By assessing the general sizes:
- \( V_D = k \frac{6Q}{d} \) is the largest.
- Next in magnitude is \( V_B = k \frac{3Q}{d} \).
- Next is \( V_C = k \frac{3\sqrt{2}Q}{d} > V_A \) (since \(3\sqrt{2} > 3\)).
- \( V_A \) will be the smallest.
Therefore, the ranking from largest to smallest is: \[ d > B > C > A \]
In symbols: \[ d > B > C > A \]
So the final ranking would be: \[ d > B > C > A \]
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