Problem 20 - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

(a) 53.33 N/C; (b) 160 N/C; (c) 480 N/C.

Step 1: Understand the Problem

We need to find the electric field at different distances from various charged objects such as a sphere, a cylinder, and a sheet. We have initial conditions for an electric field at specific distances.

Step 2: Use Gauss's Law for the Sphere

For a charged conducting sphere, the electric field outside the sphere is identical to that of a point charge located at the center. The electric field at distance \( r \) from the center of a sphere is given by \( E(r) = \frac{kQ}{r^2} \), where \( k = \frac{1}{4\pi\varepsilon_0} \). Since the field at 0.200 cm is 480 N/C, it means \( E(0.200) = \frac{kQ}{(0.200)^2} = 480 \). To find \( E(0.600) \), use the same expression: \( E(0.600) = \frac{kQ}{(0.600)^2} \).

Step 3: Calculate the Electric Field for the Sphere

The electric field falls off with the square of the distance. Knowing \( E(0.200) \) is 480 N/C, we calculate: \[ E(0.600) = 480 \left( \frac{0.200}{0.600} \right)^2 = 480 \times \left( \frac{1}{3} \right)^2 = 480 \times \frac{1}{9} = 53.33 \text{ N/C}. \]

Step 4: Use Gauss's Law for the Cylinder

For a long charged cylinder, the electric field at a distance \( r \) is given by \( E(r) = \frac{\lambda}{2\pi\varepsilon_0 r} \), where \( \lambda \) is the linear charge density. Since \( E(0.200) = 480 \), we have \( E(0.600) = \frac{\lambda}{2\pi\varepsilon_0(0.600)} \). Electric fields for cylinders decrease linearly with \( r \).

Step 5: Calculate the Electric Field for the Cylinder

Using the linear relation, \[ E(0.600) = 480 \times \frac{0.200}{0.600} = 480 \times \frac{1}{3} = 160 \text{ N/C}. \]

Step 6: Consider the Sheet of Charge Properties

For a large sheet of charge, the electric field is uniform, meaning it does not change with distance. The field is given as \( E = \frac{\sigma}{2\varepsilon_0} \), where \( \sigma \) is the surface charge density. It remains constant regardless of the distance from the sheet.

Step 7: Evaluate Electric Field from the Sheet

Since the electric field does not change with distance from the sheet, \( E(1.20) = 480 \text{ N/C}. \)

Key Concepts

Electric Field

The electric field is a vector field around charged objects. It represents the force a charge experiences in that field. Understanding how the electric field works is crucial for solving problems related to charged bodies.
When dealing with different geometries such as spheres, cylinders, and sheets, the behavior and calculation of electric fields change. The measure of the electric field depends on the charge configuration and the distance from the source of the field.
In summary, the electric field helps determine how charges interact and the strength of this interaction at any given point.

Conducting Sphere

A conducting sphere’s electric field has unique properties. When dealing with conducting spheres, all charges remain on the surface due to repulsion. This effectively makes the sphere's electric field outside mimic that of a point charge located at its center.
By applying Gauss's Law, the electric field at a distance from the center of a conducting sphere is given by the formula:

• \[ E(r) = \frac{kQ}{r^2} \]

Here, \( k \) is Coulomb's constant, \( Q \) is the total charge of the sphere, and \( r \) is the distance from the center.
Thus, the electric field decreases with the square of the distance, which is why moving further from the sphere reduces the field significantly.

Charged Cylinder

For a very long charged cylinder, the electric field behaves differently compared to a sphere. The charges create an electric field that diminishes linearly with distance. Using Gauss's Law for cylindrical symmetry, we have:

• \[ E(r) = \frac{\lambda}{2\pi\varepsilon_0 r} \]

Here, \( \lambda \) is the linear charge density, and \( \varepsilon_0 \) is the permittivity of free space.
This formula reflects the electric field diminishing as the distance from the cylinder increases. However, in contrast to a sphere, the reduction rate is proportional, not quadratic.

Uniform Sheet of Charge

The uniform sheet of charge provides an intriguing scenario for electric fields. Here, the field remains constant regardless of the distance from the sheet. This is due to the infinite plane assumption where charges distribute evenly over an extensive area
A uniform electric field is the result, expressed simply as:

• \[ E = \frac{\sigma}{2\varepsilon_0} \]

In this formula, \( \sigma \) is the surface charge density. Acting uniformly, the sheet’s field strength is independent of position, only changing with surface charge density adjustments.