Problem 22 - University Physics with Modern Physics

Question

22.22. (a) At a distance of 0.200 \(\mathrm{cm}\) from the center of a charged conducting sphere with radius \(0.100 \mathrm{cm},\) the electric field is 480 \(\mathrm{N} / \mathrm{C}\) . What is the electric field 0.600 \(\mathrm{cm}\) from the center of the sphere? (b) At a distance of 0.200 \(\mathrm{cm}\) from the axis of a very long charged conducting cylinder with radius \(0.100 \mathrm{cm},\) the electric field is 480 \(\mathrm{N} / \mathrm{C}\) . What is the electric field 0.600 \(\mathrm{cm}\) from the axis of the cylinder? (c) At a distance of 0.200 \(\mathrm{cm}\) from a large uniform sheet of charge, the electric field is 480 \(\mathrm{N} / \mathrm{C}\) . What is the electric field 1.20 \(\mathrm{cm}\) from the sheet?

Step-by-Step Solution

Answer

(a) 53.33 N/C, (b) 160 N/C, (c) 480 N/C

Step 1: Identify Known Values for Sphere

The electric field near a spherical conductor at 0.200 cm is given as 480 N/C, and the radius of the sphere is 0.100 cm.

Step 2: Apply the Electric Field Formula for Spheres

For points outside a spherical charge distribution, the electric field behaves as if all the charge were concentrated at the center. The formula is \( E = \frac{kQ}{r^2} \), where \( E \) is the electric field, \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the center.

Step 3: Calculate the Electric Field at 0.600 cm from the Sphere

The electric field at 0.600 cm will be \( E = \frac{480 \, \text{N/C} \times (0.200)^2}{(0.600)^2} = 53.33 \, \text{N/C} \).

Step 4: Identify Known Values for Cylinder

The electric field near a cylindrical conductor at 0.200 cm is given as 480 N/C, and the radius of the cylinder is 0.100 cm.

Step 5: Apply the Electric Field Formula for Cylinders

For a long line of charge, the electric field is \( E = \frac{2k\lambda}{r} \), where \( \lambda \) is the charge per unit length and \( r \) is the distance from the axis.

Step 6: Calculate the Electric Field at 0.600 cm from the Cylinder

The electric field at 0.600 cm from the axis is \( E = 480 \, \text{N/C} \times \frac{0.200}{0.600} = 160 \, \text{N/C} \).

Step 7: Identify Known Values for the Sheet

The electric field near a large charged sheet at 0.200 cm is given as 480 N/C.

Step 8: Apply the Electric Field Formula for Sheets

The electric field due to a large plane charge is constant and given by \( E = \frac{\sigma}{2\varepsilon_0} \), independent of the distance from the sheet.

Step 9: Conclusion for Sheet

The electric field at any distance, including 1.20 cm, from the charged sheet remains 480 N/C.

Key Concepts

Charged Conducting Sphere

A charged conducting sphere is an interesting element in electrostatics. Conductors distribute charge evenly across their surfaces. When examining the electric field outside such a sphere, it behaves as if all the charge is concentrated at the center of the sphere. This is because the charges redistribute uniformly over the sphere's surface.

### Electric Field Outside a Charged SphereOutside the sphere, the electric field ( E ) can be determined by Coulomb's law, and it decreases with the square of the distance from the center:

• The formula is: \( E = \frac{kQ}{r^2} \)
• Here, \( k \) is Coulomb's constant.
• \( Q \) is the total charge.
• \( r \) is the radial distance from the center of the sphere.

This relationship tells us that the field strength diminishes quickly as you move further from the sphere.

Charged Conducting Cylinder

A charged conducting cylinder, especially when it is very long, exhibits unique characteristics. Here, charges also redistribute across the surface. The electric field outside a long charged cylinder is different from that of a sphere because the geometry of the cylinder differs.

### Determining the Electric Field Outside a CylinderThe formula for the electric field ( E ) around a long charged cylinder is:

• \( E = \frac{2k\lambda}{r} \)
• Where \( \lambda \) is the charge per unit length.
• \( r \) is the distance from the axis of the cylinder.

The electric field decreases linearly, unlike the sphere where it decreases with the square of the distance.

Charged Sheet

A charged sheet represents another essential model in electrostatics, especially when discussing infinite or very large sheets. What's fascinating about a charged sheet is the behavior of its electric field.

### Unique Electric Field of a Charged SheetFor an idealized infinite charged sheet:

• The electric field is given by: \( E = \frac{\sigma}{2\varepsilon_0} \)
• \( \sigma \) denotes the surface charge density.
• \( \varepsilon_0 \) is the permittivity of free space.

An outstanding characteristic is that the electric field remains constant and is independent of the distance from the sheet. This feature stems from the idea that an infinite sheet can be considered to produce a uniform field.

Coulomb's Law

Coulomb’s Law is a foundational principle in the study of electric forces, describing how charges interact with one another. It quantifies the force between two charges. This law applies both in the macroscopic world and at atomic scales, making it versatile and powerful.

### Understanding Coulomb's LawThe law states:

• The force ( F ) between two point charges is directly proportional to the product of the magnitudes of each charge.
• It is inversely proportional to the square of the distance ( r ) between them.
• The formula is: \( F = \frac{k \cdot |q_1q_2|}{r^2} \)
• Here, \( k \) is Coulomb's constant, and \( q_1 \) and \( q_2 \) are the charges.

This law sets the groundwork for understanding how electric fields are calculated and their influences on charged objects.

Electric Field Formula

The electric field is a fundamental concept when examining how charges interact at a distance. It represents how a charge influences the space around it, described by a vector field.

### Basics of the Electric Field FormulaThe electric field ( E ) due to a single point charge is calculated as:

• \( E = \frac{F}{q} \)
• Where \( F \) is the force experienced by a small positive test charge and \( q \) is the magnitude of the test charge.
• Alternatively, for a point charge, it's also\( E = \frac{kQ}{r^2} \), aligning with Coulomb's Law.

The study of electric fields provides insight into the force fields generated by charges and helps us predict the force on another charge introduced into the field.