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title: "Two infinite slabs of uniform thickness are oriented parallel to the yz-plane. Slab 1 occupies the region \\(-2d \\le x \\le -d\\), Slab 2 occupies the region \\(0 \\le x \\le 2d\\), and the region \\(-d < x < 0\\) is empty space. The graph shows the x-component of the electric field, \\(E_x\\), as a function of position \\(x\\) along the x-axis. Which of the following statements correctly interprets the charge distribution of the system from the graph?"
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url: "https://nerd-notes.com/ubq/124542/"
date_modified: "2026-09-28T14:08:34+00:00"
---

# Two infinite slabs of uniform thickness are oriented parallel to the yz-plane. Slab 1 occupies the region \(-2d \le x \le -d\), Slab 2 occupies the region \(0 \le x \le 2d\), and the region \(-d < x < 0\) is empty space. The graph shows the x-component of the electric field, \(E_x\), as a function of position \(x\) along the x-axis. Which of the following statements correctly interprets the charge distribution of the system from the graph?

Two infinite slabs of uniform thickness are oriented parallel to the yz-plane. Slab 1 occupies the region \(-2d \le x \le -d\), Slab 2 occupies the region \(0 \le x \le 2d\), and the region \(-d < x < 0\) is empty space. The graph shows the x-component of the electric field, \(E_x\), as a function of position \(x\) along the x-axis. Which of the following statements correctly interprets the charge distribution of the system from the graph?

![A 2D Cartesian plot showing electric field component \(E_x\) on the vertical axis versus position \(x\) on the horizontal axis. Bare perpendicular axes intersect at the origin \((0, 0)\). The horizontal axis has tick marks labeled \(-2d\), \(-d\), \(0\), \(d\), and \(2d\). The vertical axis has a single positive tick mark labeled \(E_0\). A single solid curve traces the field piecewise across five intervals: for \(x < -2d\), a horizontal line segment along the horizontal axis at \(E_x = 0\); from \(x = -2d\) to \(x = -d\), a straight line segment sloping upward from \((-2d, 0)\) to \((-d, E_0)\); from \(x = -d\) to \(x = 0\), a horizontal line segment at height \(E_0\); from \(x = 0\) to \(x = 2d\), a straight line segment sloping downward from \((0, E_0)\) to \((2d, 0)\); for \(x > 2d\), a horizontal line segment along the horizontal axis at \(E_x = 0\). Vertical dashed lines extend from the tick marks at \(-2d\), \(-d\), and \(2d\) to the curve to mark boundary positions. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790604514-ktA9UW.jpg)

- **A.** The volume charge density is greatest in the region \(-d < x < 0\) because the electric field reaches its maximum constant value \(E_0\) there.
- **B.** The slab in the region \(0 < x < 2d\) carries a positive volume charge density because \(E_x > 0\) throughout that region, and the change in slope at \(x = 0\) indicates a uniform surface charge density at that boundary.
- **C.** The magnitude of the volume charge density in the slab in \(0 < x < 2d\) is twice that in the slab in \(-2d < x < -d\) because the field varies over twice the distance, and the electric potential is discontinuous at \(x = -d\).
- **D.** The slab in the region \(-2d < x < -d\) has a positive volume charge density of magnitude \(\dfrac{\varepsilon_0 E_0}{d}\), which is twice the magnitude of the negative volume charge density in the slab in \(0 < x < 2d\), and the continuity of \(E_x\) across all boundaries indicates that no surface charge resides at any interface.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/124542/*