Problem 56 - 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

(a) -1.17×10^5 N/C, outwards. (b) -8.92×10^4 N/C, outwards. (c) -1.46×10^5 N/C, outwards. Frequency of charge affects field's strength.

Step 1: Understand the problem

We have a disk charged with (-6.50 nC), and we need to find the electric field at a point P on the axis of the disk, considering three different configurations of the charge on the disk.

Step 2: Calculate electric field for uniformly charged face

The electric field for a uniformly charged disk at a point on its axis can be determined using the formula:\[ E = \frac{\sigma}{2\varepsilon_0} \left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right) \]where \(\sigma\) is the surface charge density given by \(\sigma = \frac{q}{\pi R^2}\), \(R\) is the radius of the disk, and \(z\) is the distance from the disk center to point P. Calculate \(\sigma\):\[ \sigma = \frac{-6.50 \times 10^{-9} \text{ C}}{\pi (0.0125)^2} \approx -1.32 \times 10^{-7} \text{ C/m}^2 \]Substitute values into the formula:\[ E = \frac{-1.32 \times 10^{-7}}{2 \times 8.85 \times 10^{-12}} \left(1 - \frac{0.02}{\sqrt{0.02^2 + 0.0125^2}}\right) \approx -1.17 \times 10^5 \text{ N/C} \]

Step 3: Calculate electric field for charge on rim

When all the charge is on the rim, the electric field is like that of a ring along its axis.\[ E_{\text{ring}} = \frac{k \cdot q \cdot z}{(z^2 + R^2)^{3/2}} \]Substitute the given values \(-6.50 \times 10^{-9} \text{ C}, z = 0.02 \text{ m}, R = 0.0125 \text{ m}, k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2\):\[ E = \frac{8.99 \times 10^9 \times (-6.50 \times 10^{-9}) \times 0.02}{(0.02^2 + 0.0125^2)^{3/2}} \approx -8.92 \times 10^4 \text{ N/C} \]

Step 4: Calculate electric field for point charge at center

When all the charge is at the center, it's like a point charge.\[ E = \frac{k \cdot q}{z^2} \]Substitute \(-6.50 \times 10^{-9} \text{ C} , z = 0.02 \text{ m} \):\[ E = \frac{8.99 \times 10^9 \times (-6.50 \times 10^{-9})}{0.02^2} \approx -1.46 \times 10^5 \text{ N/C} \]

Step 5: Compare the electric fields

The electric field magnitude is greatest when the charge is concentrated as a point charge at the center because there's a smaller distance for the charge distribution. It's strongest on converting to a point charge (part c), weaker when spread over the face (part a), and weakest with the charge on the rim (part b) due to increasing distance from the point P or effective charge distance from P.

Key Concepts

Surface Charge Density

When dealing with electrically charged surfaces, a key concept is surface charge density, which is denoted by the symbol \( \sigma \). It represents the amount of electrical charge per unit area on a surface. For a nonconducting disk, like in the problem above, it helps determine how the charge is spread across its surface.

To calculate surface charge density, the formula \( \sigma = \frac{q}{A} \) is used, where \( q \) is the total charge and \( A \) is the surface area. In a circular disk, the area is \( \pi R^2 \), where \( R \) is the radius of the disk. By substituting the given charge \( -6.50 \text{ nC} \) and radius \( 1.25 \text{ cm} \), we found the surface charge density to be \( -1.32 \times 10^{-7} \text{ C/m}^2 \).

Understanding surface charge density is crucial because it lays the foundation for computing electric fields produced by different charge configurations.

Point Charge

A point charge is an idealized model of a particle with a concentrated charge at a single point in space. It simplifies the complex problem of charge distribution into a more manageable one by assuming the charge is localized.

In our exercise, if the entire disk's charge is gathered at its center, it behaves as a point charge. The electric field \( E \) created by a point charge is directed radially outward (or inward if it's negative) and its magnitude is calculated by the formula \( E = \frac{k \cdot q}{r^2} \), where \( k \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance to the observation point. Here, we noticed the strongest electric field due to the minimal separation between charge and point P.

This concept helped illustrate why concentrating charge affects electric field strength significantly.

Nonconducting Disk

A nonconducting disk implies that the disk is made of a material that does not allow free movement of electrical charge across its surface or through its body. This quality impacts how charge distributes itself when placed on or distributed across the disk.

In this scenario, our disk remains charged because there is no movement of charges from one area to another. With a charge like -6.50 nC spread uniformly, it doesn't redistribute itself naturally to other areas of the disk. This characteristic allows for deeper exploration of how electric fields are formed by static charge distributions, such as those on the disk's surface and rim.

The unique properties of a nonconducting disk are fundamental to understanding the behavior of electric fields in the given configurations.

Coulomb's Law

Coulomb's Law describes the force of interaction between two point charges. This fundamental principle not only deals with force but is also crucial for calculating electric fields, which is a core part of our exercise.

In simple terms, for charges \( q_1 \) and \( q_2 \), separated by a distance \( r \), the electrostatic force \( F \) exerted by each charge on the other is \( F = \frac{k \mid q_1 q_2 \mid}{r^2} \). Here, \( k \) is Coulomb’s constant. Outside of force calculations, we can derive the electric field produced by a point charge, as explained earlier.

In the initial electric field calculations, Coulomb's Law is applied to compute how varied distributions of charge affect field intensity at given points. Its application is widespread, making it essential for understanding electric interactions both in discrete and continuous charge distributions.