Problem 88 - University Physics with Modern Physics

Question

Two very large parallel sheets are 5.00 cm apart. Sheet \(A\) carries a uniform surface charge density of \(-8.80 \space \mu\)C\(/m^2\), and sheet \(B\), which is to the right of \(A\), carries a uniform charge density of \(-11.6 \space \mu\)C\(/m^2\). Assume that the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 cm to the right of sheet \(A\); (b) 4.00 cm to the left of sheet \(A\); (c) 4.00 cm to the right of sheet \(B\).

Step-by-Step Solution

Answer

(a) 1.58 × 10⁵ N/C left; (b) 4.97 × 10⁵ N/C right; (c) 6.55 × 10⁵ N/C left.

Step 1: Understanding the Problem

We have two large sheets with charge densities \( \sigma_A = -8.80 \, \mu C/m^2 \) and \( \sigma_B = -11.6 \, \mu C/m^2 \). These sheets will produce electric fields. We need to find the net electric field at specific points with respect to these sheets.

Step 2: Electric Field Due to a Single Sheet

For an infinite sheet with uniform charge density \( \sigma \), the electric field created at a point is given by \[ E = \frac{\sigma}{2\varepsilon_0} \] where \( \varepsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \, m^2) \) is the vacuum permittivity. The field points away from the sheet if \( \sigma > 0 \) and towards the sheet if \( \sigma < 0 \).

Step 3: Electric Field from Sheet A

For sheet A with charge density \( -8.80 \, \mu C/m^2 \), we calculate the magnitude of the electric field as: \[ E_A = \frac{8.80 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \approx 4.97 \times 10^5 \, N/C \]. The direction is towards sheet A since it is negatively charged.

Step 4: Electric Field from Sheet B

Similarly, for sheet B with charge density \( -11.6 \, \mu C/m^2 \), the field is: \[ E_B = \frac{11.6 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \approx 6.55 \times 10^5 \, N/C \]. This field also points towards sheet B.

Step 5: Net Electric Field at 4.00 cm Right of Sheet A (Point a)

At this point, the field from sheet A (\(E_A\)) points left, and the effect of sheet B (\(E_B\)) is still left (being at 1 cm right of B). The net field is the difference: \[ E_{net} = E_B - E_A = 6.55 \times 10^5 - 4.97 \times 10^5 = 1.58 \times 10^5 \, N/C \] pointing left.

Step 6: Net Electric Field at 4.00 cm Left of Sheet A (Point b)

Here, only sheet A affects the field as a leftward force (towards the sheet). Sheet B doesn't affect this point. Thus, \[ E_{net} = E_A = 4.97 \times 10^5 \, N/C \] pointing right (towards sheet A).

Step 7: Net Electric Field at 4.00 cm Right of Sheet B (Point c)

In this case, both fields again direct leftward. With larger influence directly from B: \[ E_{net} = E_B = 6.55 \times 10^5 \, N/C \] pointing left (as there are no competing forces from A beyond B's right side).

Key Concepts

Surface Charge Density

Surface charge density is a measure of how much electric charge is accumulated per unit area on a surface. It is denoted by the symbol \( \sigma \) and is generally expressed in terms of microcoulombs per square meter (\( \mu C/m^2 \)). In the context of our problem, surface charge density is crucial because it determines the electric field generated by each of the sheets.
Given sheets A and B, with surface charge densities of \(-8.80 \mu C/m^2\) and \(-11.6 \mu C/m^2\) respectively, these values allow us to calculate the strength and direction of their electric fields.
It's important to note:

• A positive surface charge density implies that the electric field points away from the surface.
• A negative surface charge density implies that the electric field points towards the surface.

Understanding surface charge density helps in figuring out the influence each sheet has on the electric field at various points in the surrounding space.

Uniform Charge Distribution

Uniform charge distribution refers to charges spread evenly over a surface. When dealing with large, flat surfaces like our infinite sheets, we assume a uniform charge distribution to simplify calculations. This means that the surface charge density \( \sigma \) is constant across the surface and does not vary from one point to another.
This uniformity is key because it ensures that the electric field produced is also constant at a given distance from the surface, making calculations straightforward.
For instance, in our problem, both sheets A and B have uniform charge distributions, which means that their electric fields can be calculated using a straightforward formula for an infinite sheet. This regularity simplifies the physics involved, letting us leverage symmetry in predicting the behavior of their electric fields.

Vacuum Permittivity

Vacuum permittivity, denoted by \( \varepsilon_0 \), is a fundamental physical constant that characterizes how electric fields interact in a vacuum. Its value is approximately \( 8.85 \times 10^{-12} \, C^2/(N \, m^2) \).
This constant is critical in determining the strength of the electric field created by a charged surface. In the equation for the electric field due to an infinite sheet \( E = \frac{\sigma}{2\varepsilon_0} \), vacuum permittivity helps determine the proportionality between the surface charge density and the resulting electric field.
Understanding vacuum permittivity is important because it provides a universal scale for how electric fields propagate through space, especially in ideal, 'empty' conditions. This knowledge allows physicists and engineers to predict the behavior of electric fields around charged objects in real-world scenarios.

Infinite Sheets

Infinite sheets, in theoretical physics, are charged planes that extend indefinitely in all directions. While no real-world objects are truly infinite, this concept is used as an approximation when the dimensions of the sheets are much larger than the distance from them where the electric field is being calculated.
By treating the sheets as infinite, we can assume that the electric field generated by each sheet is uniform and can be described using simple equations derived from Gauss's Law. Such simplification is critical for solving problems like the one at hand, where the focus is on understanding the net electric field at certain points influenced by the sheets.
In practical terms, infinite sheets help us explore how electric fields behave over short distances relative to large, flat charged surfaces, making the physics more manageable and the calculations much easier.