Problem 37 - Fundamentals of Physics

Question

A square metal plate of edge length \(8.0 \mathrm{~cm}\) and negligible thickness has a total charge of \(6.0 \times 10^{-6} \mathrm{C}\). (a) Estimate the magnitude \(E\) of the electric field just off the center of the plate (at, say, a distance of \(0.50 \mathrm{~mm}\) from the center by assuming that the charge is spread uniformly over the two faces of the plate. (b) Estimate \(\bar{E}\) at a distance of \(30 \mathrm{~m}\) (large relative to the plate size) by assuming that the plate is a charged particle.

Step-by-Step Solution

Answer

(a) \(E \approx 2.65 \times 10^7 \text{ N/C}\); (b) \(\bar{E} \approx 60.0 \text{ N/C}\).

Step 1: Calculate Surface Area of the Plate

Since the plate is a square with an edge length of \(8.0 \text{ cm}\), the area \(A\) of one face is calculated as follows:\[ A = l \times l = 8.0 \text{ cm} \times 8.0 \text{ cm} = 64 \text{ cm}^2 = 64 \times 10^{-4} \text{ m}^2 = 6.4 \times 10^{-3} \text{ m}^2. \]

Step 2: Calculate Surface Charge Density

The total charge is spread over two sides, so each side has half the charge. The surface charge density \(\sigma\) is:\[ \sigma = \frac{Q}{2A} = \frac{6.0 \times 10^{-6} \text{ C}}{2 \times 6.4 \times 10^{-3} \text{ m}^2} = \frac{6.0 \times 10^{-6}}{12.8 \times 10^{-3}} = 4.6875 \times 10^{-4} \text{ C/m}^2. \]

Step 3: Calculate Electric Field just off the Plate

The magnitude of the electric field \(E\) at a point just off the center of the plate at distance \(d = 0.50 \text{ mm} = 0.50 \times 10^{-3} \text{ m}\) is given by:\[ E = \frac{\sigma}{2\varepsilon_0} = \frac{4.6875 \times 10^{-4} \text{ C/m}^2}{2 \times 8.85 \times 10^{-12} \text{ C}^2/(\text{N} \cdot \text{m}^2)} \approx 2.65 \times 10^{7} \text{ N/C}. \]

Step 4: Calculate Electric Field at Large Distance

Assume the plate is a point charge at a distance of \(30\ \text{m}\):The electric field \(\bar{E}\) is:\[ \bar{E} = \frac{kQ}{r^2} = \frac{8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \times 6.0 \times 10^{-6} \text{ C}}{(30 \text{ m})^2} = \frac{53.94 \times 10^3}{900} \approx 60.0 \text{ N/C}. \]

Key Concepts

Surface Charge Density

Surface charge density, denoted by the Greek letter \( \sigma \), is a measure of how much electric charge is accumulated on a surface. It is calculated as the charge per unit area. When dealing with a charged plate, like in the exercise mentioned, you distribute the total charge over the surface. Here’s how it can be understood:

• Imagine the plate as having two faces, and the total charge is equally spread over these faces.
• The formula for surface charge density is \( \sigma = \frac{Q}{2A} \), where \( Q \) is the total charge and \( A \) is the area of one side of the plate.
• In the problem, we calculated \( \sigma \) as \( 4.6875 \times 10^{-4} \text{ C/m}^2 \).

Understanding surface charge density is crucial because it helps in determining the electric field produced by the surface
which affects how the charge interacts with its surroundings.

Electric Field Calculation at a Point

The electric field calculation at a specific point depends on several factors, such as where the point is located relative to the charge distribution. In the exercise, we needed to find the electric field just off the center
of a charged plate.

• The electric field \( E \) near a uniformly charged plate is calculated considering the charge distribution.
• For points very close to the plate, we use the equation \( E = \frac{\sigma}{2\varepsilon_0} \), where \( \varepsilon_0 \) is the electric permittivity of free space.
• This formula assumes an infinite plane model, providing a good approximation near the center of large plates
  and helps in calculating \( E \) effectively for close distances.

With this formula, we find the field strength close to the plate, which was approximated as \( 2.65 \times 10^7 \text{ N/C} \) in the problem, indicating a strong electric field due to the surface charge density.

Charged Plate

A charged plate can be thought of as a flat conductor where electric charge is spread across its surface. It can serve as a foundational model for understanding electric fields. The behavior of the electric field generated by a charged plate differs based on the distance:

• Close to the plate, we consider the charge distribution uniform, utilizing specific equations for better approximation.
• At a distance (much larger than the dimensions of the plate), the field can be approximated by considering the plate as a point charge. This simplification helps when the observer is far away, making calculation easier.
• In the example, for a distance of 30 meters, the charged plate's field was treated as originating from a point charge, resulting in\( \bar{E} \approx 60.0 \text{ N/C} \).

Understanding these characteristics is essential in physics as it aids in solving problems related to charge distributions and their resultant fields under various conditions.