---
title: "In a region of the \\(xy\\)-plane, the electric field is given by \\(\\vec{E} = \\alpha x\\,\\hat{i}\\), where \\(\\alpha\\) is a positive constant. Point \\(A\\) is located at \\((x_0, 0)\\) and point \\(B\\) is located at \\((2x_0, y_0)\\), where \\(x_0 > 0\\) and \\(y_0 > 0\\).  Which of the following statements correctly evaluates the electric potential difference \\(V_B – V_A\\) and provides the correct physical justification?"
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url: "https://nerd-notes.com/ubq/121080/"
date_modified: "2026-08-23T04:57:42+00:00"
---

# In a region of the \(xy\)-plane, the electric field is given by \(\vec{E} = \alpha x\,\hat{i}\), where \(\alpha\) is a positive constant. Point \(A\) is located at \((x_0, 0)\) and point \(B\) is located at \((2x_0, y_0)\), where \(x_0 > 0\) and \(y_0 > 0\).

Which of the following statements correctly evaluates the electric potential difference \(V_B – V_A\) and provides the correct physical justification?

In a region of the \(xy\)-plane, the electric field is given by \(\vec{E} = \alpha x\,\hat{i}\), where \(\alpha\) is a positive constant. Point \(A\) is located at \((x_0, 0)\) and point \(B\) is located at \((2x_0, y_0)\), where \(x_0 > 0\) and \(y_0 > 0\).

Which of the following statements correctly evaluates the electric potential difference \(V_B - V_A\) and provides the correct physical justification?

![A two-dimensional Cartesian coordinate system with a horizontal axis labeled x and a vertical axis labeled y intersecting at an origin labeled O. On the positive x-axis, two tick marks are labeled x_0 and 2x_0. On the positive y-axis, one tick mark is labeled y_0. A solid dot on the x-axis at x_0 is labeled A. A solid dot in the first quadrant at the intersection of a horizontal dashed line from y_0 and a vertical dashed line from 2x_0 is labeled B. Four horizontal right-pointing arrows representing the electric field are shown in the first quadrant: two shorter arrows at x = x_0 and two longer arrows at x = 2x_0. A label \vec{E} is located near the arrows. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787461062-S0H5J2.jpg)

- **A.** The electric potential difference \(V_B - V_A\) is equal to \(+\dfrac{3}{2}\alpha x_0^2\) because the electric field points in the \(+x\)-direction, which causes the electric potential to increase with increasing \(x\).
- **B.** The electric potential difference \(V_B - V_A\) is equal to \(-\dfrac{3}{2}\alpha x_0^2 - \alpha x_0 y_0\) because the line integral of the electric field depends directly on the total diagonal path length connecting point \(A\) and point \(B\).
- **C.** The electric potential difference \(V_B - V_A\) is equal to \(+\dfrac{3}{2}\alpha x_0^2\) because the electrostatic field does positive work on a positive test charge moving from \(A\) to \(B\).
- **D.** The electric potential difference \(V_B - V_A\) is equal to \(-\dfrac{3}{2}\alpha x_0^2\) because the electric field has no component parallel to the \(y\)-axis, so displacement in the \(y\)-direction contributes zero to the line integral \(\int \vec{E}\cdot d\vec{\ell}\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/121080/*