Question

Rank the electric potentials at the four points shown in the figure below from largest to smallest. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.)

Incorrect: Your answer is incorrect.

A square of side d has positive charge Q at its bottom left corner and positive charge 2Q at its bottom right corner. Four points are labeled as follows.
Point A is at the top left corner.
Point B is at the top right corner.
Point C is at the center of the square.
Point D is centered on the bottom edge of the square.

Answer 1

To determine the electric potentials at the four points due to the charges, we need to consider how electric potential (V) due to a point charge is defined:

\[
V = k \frac{Q}{r}
\]

Where:
- \(V\) is the electric potential.
- \(k\) is Coulomb's constant.
- \(Q\) is the charge.
- \(r\) is the distance from the charge to the point of interest.

### Analyzing Each Point:

1. **Point A (Top Left Corner)**:
- Distance from charge \(Q\) (bottom left corner) is \(d\).
- Distance from charge \(2Q\) (bottom right corner) is \(\sqrt{2}d\).
- Potential \(V_A = k \frac{Q}{d} + k \frac{2Q}{\sqrt{2}d} = k\frac{Q}{d} + k\frac{2Q}{\sqrt{2}d}\).

2. **Point B (Top Right Corner)**:
- Distance from charge \(Q\) (bottom left corner) is \(\sqrt{2}d\).
- Distance from charge \(2Q\) (bottom right corner) is \(d\).
- Potential \(V_B = k \frac{Q}{\sqrt{2}d} + k \frac{2Q}{d} = k\frac{Q}{\sqrt{2}d} + k\frac{2Q}{d}\).

3. **Point C (Center of Square)**:
- Distance from charge \(Q\) (bottom left corner) is \(\frac{\sqrt{2}}{2}d\).
- Distance from charge \(2Q\) (bottom right corner) is also \(\frac{\sqrt{2}}{2}d\).
- Potential \(V_C = k \frac{Q}{\frac{\sqrt{2}}{2}d} + k \frac{2Q}{\frac{\sqrt{2}}{2}d} = k\frac{Q \sqrt{2}}{d} + k\frac{2Q \sqrt{2}}{d} = k\frac{3Q\sqrt{2}}{d}\).

4. **Point D (Centered on Bottom Edge)**:
- Distance from charge \(Q\) (bottom left corner) is \(\frac{d}{2}\).
- Distance from charge \(2Q\) (bottom right corner) is also \(\frac{d}{2}\).
- Potential \(V_D = k \frac{Q}{\frac{d}{2}} + k \frac{2Q}{\frac{d}{2}} = k\frac{2Q}{d} + k\frac{4Q}{d} = k\frac{6Q}{d}\).

### Summary of Potentials:
- \(V_A = k \left( \frac{Q}{d} + \frac{2Q}{\sqrt{2}d} \right)\)
- \(V_B = k \left( \frac{Q}{\sqrt{2}d} + \frac{2Q}{d} \right)\)
- \(V_C = k \frac{3Q\sqrt{2}}{d}\)
- \(V_D = k \frac{6Q}{d}\)

### Comparison:
- \(V_D\) is the largest due to the maximum contributions from both charges.
- \(V_C\) follows, as it has the sum of potentials from both charges at equal distances.
- \(V_A\) and \(V_B\) are more complicated; however, generally we can compare:
- Since both \(Q\) and \(2Q\) contribute to \(V_A\) and \(V_B\) but differ in distance to the charges, \(V_A < V_B\).

Thus, we can rank the potentials from largest to smallest as follows:

\[
V_D > V_C > V_B > V_A
\]