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The problem involves calculating the electric potential difference (\(\Delta V\)) between two points, \(A\) and \(B\), in the presence of a uniform electric field (\(E = 1100 \, \text{N/C}\)) pointing in the negative \(x\)-direction.

### Key Information:
- The electric field points in the negative \(x\)-direction.
- The coordinates of the points:
  - \(A\) is located at (0, 0).
  - \(B\) is located at (4.0 cm, 0).
- You are asked to find the potential difference between points \(A\) and \(B\).

### Formula:
The electric potential difference between two points in a uniform electric field is given by:
\[
\Delta V = V_B - V_A = -E \cdot d \cdot \cos(\theta)
\]
Where:
- \(E = 1100 \, \text{N/C}\) (electric field strength),
- \(d = 4.0 \, \text{cm} = 0.04 \, \text{m}\) (distance between \(A\) and \(B\)),
- \(\theta = 0^\circ\) (since the electric field is along the \(x\)-axis and the distance between \(A\) and \(B\) is also along the \(x\)-axis).

### Solution:
The angle \(\theta = 0^\circ\) means \(\cos(0^\circ) = 1\), so the potential difference becomes:
\[
\Delta V = -E \cdot d = -(1100) \cdot (0.04) = -44 \, \text{V}
\]

Therefore, the correct answer for the potential difference \(\Delta V\) between points \(A\) and \(B\) is:

\[
\Delta V = -44 \, \text{V}
\]

Would you like further clarification on this? 

Here are five related questions you might consider:
1. How would the potential difference change if the electric field direction were reversed?
2. What would the potential difference be if the distance between \(A\) and \(B\) were doubled?
3. How does the angle between the electric field and the displacement affect the potential difference?
4. What would happen to the potential difference if the electric field strength were halved?
5. How does this concept relate to real-world applications of electric fields, such as in capacitors?

**Tip**: Always make sure the direction of the electric field and the position of the points are consistent before calculating the potential difference!

What is the difference in electric potential, ΔV = VB - VA, between points A and B in the presence of a uniform electric field of 1100 N/C pointing in the negative x-direction?

Learn how to calculate the electric potential difference between points A and B in a uniform electric field of 1100 N/C. This problem involves key physics concepts such as the electric field, trigonometry, and the formula for potential difference.

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