Problem 55 - University Physics with Modern Physics

Question

A charge of \(-6.50 \mathrm{nC}\) is spread uniformly over the surface of one face of a nonconducting disk of radius 1.25 \(\mathrm{cm}\) . (a) Find the magnitude and direction of the electric field this disk produces at a point \(P\) on the axis of the disk a distance of 2.00 \(\mathrm{cm}\) from its center. (b) Suppose that the charge were all pushed away from the center and distributed uniformly on the outer rim of the disk. Find the magnitude and direction of the electric field at point \(P\) . (c) If the charge is all brought to the center of the disk, find the magnitude and direction of the electric field at point \(P .\) (d) Why is the field in part (a) stronger than the field in part (b)? Why is the field in part (c) the strongest of the three fields?

Step-by-Step Solution

Answer

The strongest field is when the charge is at the center, weakest for the outer rim. Uniform field is stronger than the rim but weaker than the center.

Step 1: Identify the Problem

We are asked to find the magnitude and direction of the electric field produced by a charge distributed on a disk under three different distribution conditions: uniformly spread over the surface, spread on the outer rim, and all concentrated at the center.

Step 2: Calculate Electric Field (Uniform Surface Distribution)

For part (a), use the formula for the electric field due to a uniformly charged disk:\[ E = \frac{\sigma}{2\varepsilon_0} \left( 1 - \frac{z}{\sqrt{z^2 + R^2}} \right) \]where \(\sigma = \frac{Q}{\pi R^2}\) is the surface charge density, \(\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2\), \(z = 2.00 \, \text{cm}\), and \(R = 1.25 \, \text{cm}\). Plug in the values, and solve for \(E\).

Step 3: Evaluate Electric Field (Charge at Outer Rim)

For part (b), consider the effect of the charge distributed along the rim. The field is more complicated to calculate exactly without integration, but generally, the effective result is weaker compared to when the charge is spread uniformly, as each piece of charge is now further from point \(P\). This implies that the calculated field will be less than in part (a).

Step 4: Calculate Electric Field (Charge at Center)

For part (c), if all the charge is at the center, use the formula for the electric field of a point charge:\[ E = \frac{kQ}{z^2} \]where \(k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2\) and \(z = 0.02 \, \text{m}\). This results in the strongest field compared to the other configurations.

Step 5: Analyze Differences in Field Strength

For part (d), since the field for a point charge (all charge at the center) depends inversely on \(z^2\), more charge density at one point results in the strongest field. In part (b), the field is weaker due to the distribution around the rim, increasing the effective distance from point \(P\). Part (a) shows a strong field but less than part (c), due to partial distribution effect; however, it is more centralized than part (b).

Key Concepts

Charge Distribution

Charge distribution describes how electric charge is arranged over a given area or volume. In our exercise, electric charge is distributed across a nonconducting disk in different ways.

• In part (a), the charge is spread uniformly across the surface of the disk.
• In part (b), the same charge is located on the rim of the disk.
• In part (c), all the charge is concentrated at the center of the disk.

Each distribution affects the electric field created by the disk differently. Uniform distribution across the disk offers a balanced electric field, while concentrating the charge at a point, like the center in part (c), often results in the strongest electric field at specific points away from the charge.

Nonconducting Disk

A nonconducting disk is an insulating surface that does not allow electric charges to move freely across it. This setup is crucial because it maintains the charge distribution without any influence from external factors. When dealing with the nonconducting surface in this problem:

• Charges remain fixed at their allocated locations.
• The disk's insulating nature ensures that the charge does not spread or redistribute under any external electric field influence.

The nonconducting property thereby allows us to accurately calculate and compare the electric fields generated in each of the given scenarios.

Surface Charge Density

Surface charge density, denoted by \( \sigma \), is the amount of electric charge per unit area on a surface, often simplified by the formula \( \sigma = \frac{Q}{\pi R^2} \), where \( Q \) is the total charge and \( R \) is the radius of the disk. In case of our exercise:

• The charge density is calculated based on the charge distribution over the area of the disk surface.
• A uniformly distributed charge across a nonconducting disk will have a constant surface charge density, which directly influences the overall strength of the electric field.

Understanding surface charge density is important because it helps determine how strong the electric field will be at various points in relation to the disk.

Electric Field Magnitude and Direction

The electric field is a vector quantity, described by both its magnitude and direction, representing the force a charge would experience per unit of charge. In the context of the exercise, evaluating the electric field at a point \( P \) along the axis involves:

• For uniform charge distribution (part a), using the formula: \[ E = \frac{\sigma}{2\varepsilon_0} \left( 1 - \frac{z}{\sqrt{z^2 + R^2}} \right) \] to determine that the field is directed away from the disk surface if the disk is positively charged, or towards the disk if negatively charged.
• For rim distribution (part b), acknowledging a weaker field due to the increased average distance from point \( P \).
• For point charge at the center (part c), using the point charge formula: \[ E = \frac{kQ}{z^2} \] to establish that the electric field is the strongest, with the direction depending on the sign of the charge.

Determining the magnitude and direction allows us to visualize how the electric field behaves under different charge distributions on the disk.