Problem 39 - University Physics with Modern Physics

Question

The electric field at the surface of a charged, solid, copper sphere with radius 0.200 m is 3800 N\(/\)C, directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?

Step-by-Step Solution

Answer

The potential at the center of the sphere is the same as the potential at the surface, calculated using the sphere's charge.

Step 1: Identify the Relationship Between Electric Field and Charge

The electric field \( E \) at the surface of a charged sphere is linked to the charge \( Q \) with the formula \( E = \frac{Q}{4\pi\varepsilon_0 r^2} \), where \( r \) is the radius of the sphere and \( \varepsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2\)). The given electric field \( E \) is 3800 N/C and the radius \( r \) is 0.200 m.

Step 2: Solve for the Charge on the Sphere

Rearrange the equation to solve for \( Q \): \( Q = E \cdot 4\pi\varepsilon_0 r^2\). Substitute the known values into the equation: \( Q = 3800 \cdot 4\pi\cdot 8.85 \times 10^{-12} \cdot (0.200)^2 \). Calculate this to find \( Q \).

Step 3: Understanding Potential at the Sphere's Surface

The electric potential \( V \) at the surface of the sphere due to its own charge is determined using the formula \( V = \frac{Q}{4\pi\varepsilon_0 r} \). This uses the charge \( Q \) calculated in Step 2 and the radius \( r = 0.200 \text{ m} \).

Step 4: Calculate the Potential at the Sphere's Surface

Substitute the charge \( Q \) found in Step 2 into the formula for potential: \( V = \frac{Q}{4\pi\varepsilon_0 \cdot 0.200} \). Compute the result to find the potential at the surface.

Step 5: Determine the Potential at the Sphere's Center

For a conductor such as a copper sphere, the electric potential is constant throughout its volume. Therefore, the potential at the center of the sphere is the same as the potential at the surface.

Key Concepts

Electric Field

The electric field is a fundamental concept in electromagnetism. It represents the force per charge that would be exerted on any test charge placed in the vicinity of another charged object. In this exercise, we're considering a charged sphere. The electric field at the surface of the sphere is calculated using the formula:

• \( E = \frac{Q}{4\pi\varepsilon_0 r^2} \)

Here, \( E \) is the electric field, \( Q \) is the charge on the sphere, \( r \) is the radius of the sphere, and \( \varepsilon_0 \) represents the permittivity of free space. When we say the electric field is directed toward the center, it indicates that the sphere is negatively charged.
A negative electric field suggests that positive charges outside of the sphere would be attracted towards it. Understanding the electric field is crucial, as it helps us analyze how the force acts over space due to the presence of the charged sphere.

Charged Sphere

A charged sphere, in this context, is a solid body with a symmetrical charge distribution. Due to its spherical shape, it has a uniform electric field on its surface, given by the formula mentioned earlier. This uniform field allows us to utilize Gauss's Law, which simplifies calculations regarding charge and electric field.

For a conductor, like our copper sphere, the charge distributes uniformly on its surface. This occurs because charges repel equivalent charges and tend to move as far away from each other as possible, which is accomplished by residing on the surface.

Some important facts about charged spheres include:

• In conductors, the electric field inside is zero.
• The potential at any point inside the sphere is constant and equal to the potential on the surface.

Knowing this allows us to directly determine that the potential throughout a conducting sphere is constant, as long as we have calculated the potential at the surface correctly.

Permittivity of Free Space

The permittivity of free space, denoted as \( \varepsilon_0 \), is a fundamental constant in electromagnetism. It characterizes how electric fields interact with the vacuum of free space. In the context of this exercise, \( \varepsilon_0 \) appears in the formula for electric field and electric potential:

• \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \).

This constant represents the ability of free space to "permit" electric field lines and affects the capacitance of objects in a vacuum. With charged objects such as our sphere, the permittivity is crucial in determining how much charge can exist without resulting in overly strong electric fields.
The concept of permittivity is essential in many areas of physics and engineering, particularly when calculating the strength of electric fields and forces between charges. By understanding \( \varepsilon_0 \), students can better comprehend why certain configurations of charge produce the fields and potentials they do. This foundational concept is a building block for more complex studies in electrostatics and electromagnetism.