Problem 80 - Fundamentals of Physics

Question

Charge of uniform surface density \(8.00 \mathrm{nC} / \mathrm{m}^{2}\) is distributed over an entire \(x y\) plane; charge of uniform surface density \(3.00 \mathrm{nC} / \mathrm{m}^{2}\) is distributed over the parallel plane defined by \(z=2.00 \mathrm{~m}\). Determine the magnitude of the electric field at any point having a \(z\) coordinate of (a) \(1.00 \mathrm{~m}\) and (b) \(3.00 \mathrm{~m}\).

Step-by-Step Solution

Answer

(a) 282.49 N/C upward, (b) 621.47 N/C upward.

Step 1: Understanding the Problem

We have two charged planes with uniform surface densities. The first plane is on the xy-plane, with a surface density of \(8.00 \, \text{nC/m}^2\), and the second is parallel to it at \(z = 2.00 \, \text{m}\), with a surface density of \(3.00 \, \text{nC/m}^2\). We will determine the electric field at \(z = 1.00 \, \text{m}\) and \(z = 3.00 \, \text{m}\).

Step 2: Electric Field Due to a Single Plane of Charge

The electric field due to a main plane with surface charge density \(\sigma\) is given by \(E = \frac{\sigma}{2\varepsilon_0}\), directed away from the plane if \(\sigma\) is positive and towards it if \(\sigma\) is negative, where \(\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2\).

Step 3: Electric Field at z = 1.00 m

At \(z = 1.00\, \text{m}\) (between the two planes), the electric field due to the lower plane (\(\sigma_1 = 8.00 \, \text{nC/m}^2 = 8.00\times10^{-9} \, \text{C/m}^2\)) points upward and is calculated as \(E_1 = \frac{8.00\times10^{-9}}{2\times 8.85\times10^{-12}} = 451.98 \, \text{N/C}\). The electric field due to the upper plane (\(\sigma_2 = 3.00 \, \text{nC/m}^2 = 3.00\times10^{-9} \, \text{C/m}^2\)) points downward and is calculated as \(E_2 = \frac{3.00\times10^{-9}}{2\times 8.85\times10^{-12}} = 169.49 \, \text{N/C}\). The net electric field is \(E = E_1 - E_2 = 451.98 \, \text{N/C} - 169.49 \, \text{N/C} = 282.49 \, \text{N/C}\) upward.

Step 4: Electric Field at z = 3.00 m

At \(z = 3.00\, \text{m}\) (above both planes), the electric field due to the lower plane still points upward with the same magnitude \(E_1 = 451.98 \, \text{N/C}\). The field due to the upper plane now also points upward since \(z = 3.00\, \text{m}\) is above it, \(E_2 = 169.49 \, \text{N/C}\). The net electric field becomes \(E = E_1 + E_2 = 451.98 \, \text{N/C} + 169.49 \, \text{N/C} = 621.47 \, \text{N/C}\) upward.

Key Concepts

Uniform Surface Charge Density

Uniform surface charge density refers to the distribution of electric charge over a surface, where the charge per unit area remains constant across the entire surface. Imagine spreading peanut butter evenly over a slice of bread. Every part of the bread has the same amount of peanut butter per unit area, just like our charged planes have the same amount of charge per square meter.

In the given problem, we have two charged planes:

• The first plane has a charge density of \(8.00 \, \text{nC/m}^2\) and is located in the \(xy\)-plane.
• The second plane, parallel to the first one, has a charge density of \(3.00 \, \text{nC/m}^2\) and is positioned at \(z = 2.00 \, \text{m}\).

Even though these two planes have different charge densities, each plane itself has a uniform surface charge distribution. This uniformity simplifies calculations of the electric field, as we can use straightforward formulas without worrying about variations in the charge distribution across the planes.

When calculating the electric field caused by a uniform surface charge density, the value of the surface charge density \(\sigma\) directly determines the strength of the electric field.

Parallel Charged Planes

Parallel charged planes are a classic scenario often encountered in electrostatics. In the problem, we have two parallel planes that are charged uniformly, meaning the charges are spread evenly across each plane.

To visualize parallel charged planes, think of them as two sheets hanging parallel to each other, with one above the other. The electric fields they create interact in interesting ways, especially when determining the net electric field at different points. The position of the point between or outside these planes significantly affects the resulting electric field.

In particular:

• When you're between the planes, the electric fields from both planes partially cancel each other out if they are in opposite directions. This is because, for each plane, the electric field points directly away from a positive charge or towards a negative charge.
• If \(z = 1.00 \, \text{m}\), which is between the planes, the field from the lower plane (positive direction) and the field from the upper plane effectively subtract due to their opposing directions.
• When above both planes, as at \(z = 3.00 \, \text{m}\), fields from both planes point in the positive direction, adding together to create a stronger field upward.

The interplay between the fields makes the study of parallel charged planes not only a fundamental aspect of electric fields but also a compelling exercise in understanding how fields combine.

Gauss's Law

Gauss's Law is a powerful tool used to calculate electric fields, particularly when dealing with symmetrical charge distributions like planes. It states that the electric flux through a closed surface is proportional to the charge enclosed by that surface.

Mathematically, Gauss's Law is expressed as:\[\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{ ext{enclosed}}}{\varepsilon_0}\]where:

• \(\Phi_E\) is the electric flux through a closed surface.
• \(\vec{E}\) is the electric field.
• \(Q_{\text{enclosed}}\) is the total charge enclosed within the surface.
• \(\varepsilon_0\), the permittivity of free space, is a constant, approximately \(8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2\).

In our problem, each plane acts as a part of its own Gaussian surface. This is why we can use the formula \(E = \frac{\sigma}{2\varepsilon_0}\) for the electric field generated by each plane on its own. The symmetry and uniform charge distribution mean the solutions remain manageable even for higher complexities. By understanding Gauss’s Law, students can see how elegant mathematics can simplify seemingly complicated physical scenarios.