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 N/C, (b) 622 N/C.

Step 1: Identify the Problem

We need to find the electric field at two specific points due to two charged planes with uniform surface charge densities of \(8.00 \text{ nC/m}^2\) for the plane at \(z = 0\), and \(3.00 \text{ nC/m}^2\) for the plane at \(z = 2.00\, \text{m}\). We will use the principle of superposition for electric fields.

Step 2: Use Gauss's Law

The electric field \(E\) due to an infinite plane with surface charge density \(\sigma\) is given by Gauss's law as \(E = \frac{\sigma}{2\varepsilon_0}\), where \(\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/ \text{N} \, \text{m}^2\) is the permittivity of free space.

Step 3: Calculate the Electric Field from each plane

For the plane at \(z=0\):\[E_1 = \frac{8.00 \times 10^{-9} \text{ C/m}^2}{2 \times 8.85 \times 10^{-12} \text{ C}^2/ \text{N} \, \text{m}^2} \approx 452 \text{ N/C}\]For the plane at \(z=2\, \text{m}\):\[E_2 = \frac{3.00 \times 10^{-9} \text{ C/m}^2}{2 \times 8.85 \times 10^{-12} \text{ C}^2/ \text{N} \, \text{m}^2} \approx 170 \text{ N/C}\]

Step 4: Determine the net electric field at \(z=1\, \text{m}\)

At \(z=1\, \text{m}\), the point is between the two charged planes. The electric field from the \(z=0\) plane pushes away (in the positive z-direction), and from the \(z=2\, \text{m}\) plane, it also pushes away (in the negative z-direction). Both fields are additive in opposite directions:\[E_{\text{net}} = E_1 - E_2 = 452 \text{ N/C} - 170 \text{ N/C} = 282 \text{ N/C}\] (direction is towards the plane with larger \(\sigma\)).

Step 5: Determine the net electric field at \(z=3\, \text{m}\)

At \(z=3\, \text{m}\), the point is above both planes. The electric field from \(z=0\) is in the positive z-direction, whilst from \(z=2\, \text{m}\) is also in the positive z-direction because the point is above:\[E_{\text{net}} = E_1 + E_2 = 452 \text{ N/C} + 170 \text{ N/C} = 622 \text{ N/C}\] (direction is by convention positive)

Step 6: Finalize the solution

For \(z=1\, \text{m}\), the electric field is \(282 \text{ N/C}\) directed towards the \(z=0\) plane. For \(z=3\, \text{m}\), the net electric field is \(622 \text{ N/C}\) pointing upwards in the positive z-direction.

Key Concepts

Gauss's Law

Gauss's Law is a fundamental principle that helps us determine the electric field generated by symmetrical charge distributions like infinite planes, lines, or spheres. The law states that the electric field flux through any closed surface is proportional to the total charge enclosed within that surface.

In mathematical terms, Gauss's Law is expressed as: \( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_\text{enc}}{\varepsilon_0} \). This equation tells us that the integral of the electric field \( \mathbf{E} \) over a closed surface (represented by \( d\mathbf{A} \)) is equal to the enclosed charge \( Q_\text{enc} \) divided by the permittivity of free space \( \varepsilon_0 \).

For an infinite plane with uniform surface charge density \( \sigma \), Gauss's Law simplifies to \( E = \frac{\sigma}{2\varepsilon_0} \). This simplification occurs because the electric field is perpendicular to the surface and has the same magnitude on both sides of the plane, allowing us to consider a Gaussian "pillbox" that simplifies calculation. This result is particularly useful when analyzing fields between, above, or below infinite planar surfaces.

Surface Charge Density

Surface charge density, denoted as \( \sigma \), is the amount of electric charge per unit area on a surface. It is measured in units of coulombs per square meter (C/m²).

In practical scenarios like our exercise, we have two separate planes, each with a different surface charge density. For instance:

• The plane at \( z = 0 \) has a surface charge density of \( 8.00 \text{ nC/m}^2 \).
• The plane at \( z = 2.00 \text{ m} \), a bit offset in space, has a surface charge density of \( 3.00 \text{ nC/m}^2 \).

These values directly influence the electric fields these planes produce due to their uniform charge distribution.

The greater the surface charge density, the stronger the electric field produced. This stems from Gauss's Law, where higher \( \sigma \) values result in higher electric field magnitudes \( E \). Thus, knowing \( \sigma \) helps us predict the strength and direction of the electric field a charged surface will impart onto its surroundings.

Superposition Principle

The superposition principle allows us to calculate the resultant electric field when multiple sources contribute to it. According to this principle, the net electric field at any point is the vector sum of the electric fields due to the individual charges or charged surfaces.

In the scenario we are discussing, there are two charged planes, one at \( z = 0 \) and another at \( z = 2.00 \, \text{m} \). Each plane creates its electric field, denoted \( E_1 \) and \( E_2 \), respectively:

• For the plane at \( z = 0 \): \( E_1 = 452 \text{ N/C} \).
• For the plane at \( z = 2.00 \text{ m}: \) \( E_2 = 170 \text{ N/C} \).

To find the total electric field at a point located between or outside these planes, we simply add the contributions vectorially.

At \( z = 1.00 \text{ m} \), located between the planes, the electric fields oppose each other. So, \( E_{\text{net}} = E_1 - E_2 = 282 \text{ N/C} \). However, at \( z = 3.00 \text{ m} \), where both fields push in the same direction, \( E_{\text{net}} = E_1 + E_2 = 622 \text{ N/C} \). This principle underscores the ability to handle complex electric field scenarios by breaking them down into simpler parts.