Problem 32 - University Physics with Modern Physics

Question

A total electric charge of 3.50 \(\mathrm{nC}\) is distributed uniformly over the surface of a metal sphere with a radius of \(24.0 \mathrm{cm} .\) If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) \(48.0 \mathrm{cm} ;(\mathrm{b}) 24.0 \mathrm{cm}\) (c) \(12.0 \mathrm{cm} .\)

Step-by-Step Solution

Answer

(a) 65.47 V, (b) 130.9 V, (c) 130.9 V.

Step 1: Understanding Potential at a Distance from a Charged Sphere

The potential (V) due to a charged sphere is calculated using the formula:\[ V = \frac{kQ}{r} \]where \( k \) is Coulomb's constant \( (8.99 \times 10^9 \ \text{N m}^2/\text{C}^2) \), \( Q \) is the total charge, and \( r \) is the distance from the center of the sphere. Note that if you are outside the sphere, the sphere can be treated as a point charge situated at its center for calculating potential.

Step 2: Calculate the Potential at 48.0 cm (Outside the Sphere)

The distance, \( r = 48.0 \ \text{cm} = 0.48 \ \text{m} \).Using the formula, substitute the known values:\[ V_{48} = \frac{8.99 \times 10^9 \times 3.50 \times 10^{-9}}{0.48} \]Calculate to find \( V_{48} \).

Step 3: Calculate the Potential at 24.0 cm (On the Surface of the Sphere)

For the potential on the surface, \( r = 24.0 \ \text{cm} = 0.24 \ \text{m} \).Using the formula:\[ V_{24} = \frac{8.99 \times 10^9 \times 3.50 \times 10^{-9}}{0.24} \]Calculate to find \( V_{24} \).

Step 4: Understand that Potential is Constant Inside the Sphere

The potential inside a conductor with a uniform charge distribution is constant and equals the potential on its surface. Thus, the potential at \( r = 12.0 \ \text{cm} \) is the same as at \( r = 24.0 \ \text{cm} \); \( V_{12} = V_{24} \).

Step 5: Calculate Numerical Values

Calculate the potentials using the formula:1. For \( V_{48} \): \[ V_{48} = \frac{8.99 \times 10^9 \times 3.50 \times 10^{-9}}{0.48} \approx 65.47 \ \text{V} \]2. For \( V_{24} \) and \( V_{12} \): \[ V_{24} = \frac{8.99 \times 10^9 \times 3.50 \times 10^{-9}}{0.24} \approx 130.9 \ \text{V} \] Therefore, \( V_{12} = 130.9 \ \text{V} \).

Key Concepts

Charged Sphere

A charged sphere is a fascinating concept in physics where the entire charge is uniformly distributed over the surface, typically of a metal sphere. In such cases, due to symmetry and uniformity, we can treat the sphere as if all its charge were concentrated at its center for the purpose of calculating the electric potential or field. This simplification is valid as long as the point where we are calculating the potential is outside the sphere.

In our exercise, the sphere has a total charge of 3.50 nC (nanocoulombs), and we look into the potential at different distances from its center. Whenever we are outside the sphere, we apply the formula of a point charge to find the potential. This is one of the key advantages of a charged sphere in electric potential problems, making calculations much simpler and straightforward.

Coulomb's Law

Coulomb's Law is fundamental in understanding electric forces and potential. It expresses the force between two point charges and is given by the formula:\[ F = \frac{k |q_1 q_2|}{r^2} \]where:

• \( F \) is the force between the charges,
• \( k \) is Coulomb's constant \((8.99 \times 10^9 \ \text{N m}^2/\text{C}^2)\),
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
• \( r \) is the distance between the charges.

Instead of focusing on the force, when calculating the electric potential, which is our interest here, we use a similar concept but slightly adjust the formula. The potential \( V \) at a distance \( r \) due to a point charge \( Q \) is given by:\[ V = \frac{kQ}{r} \] This is directly utilized when working with charged spheres. Remember, for a charged sphere, if you are calculating the potential outside, \( r \) would be the distance from the center to the point where potential is measured.

Uniform Charge Distribution

Uniform charge distribution refers to the charge spread evenly across the sphere's surface, which is crucial for simplifying calculations. This uniform spread means that every part of the sphere's surface has the same amount of charge per unit area.

Because of this symmetry:

• Calculations become simpler – the distribution allows us to treat the sphere as a point charge when considering points outside.
• Potential calculations inside the sphere become straightforward as well, since the entire interior is at the same potential as the surface.

This uniformity underlies why it's possible to claim that the potential inside remains constant, equal to the surface potential. The uniform charge distribution makes these types of problems easier to solve, regardless of whether the point of interest is inside or outside the sphere.

Potential Inside a Conductor

The potential inside a conductor that holds a static charge is a concept driven by electrostatic equilibrium. In such equilibrium, charges arrange themselves on the surface, ensuring that the electric field inside the conductor is zero. This absence of an electric field inside the conductor means there is no difference in potential throughout its interior.

Hence:

• The potential inside equals the potential on the surface.
• For the charged sphere, at \( r = 12.0 \text{ cm} \), the potential remains the same as that at \( r = 24.0 \text{ cm} \), which is the surface.
• This results from the conductor's properties that negate potential differences internally.

It's this unique behavior of conductors in electrostatic conditions that guarantees a constant potential within. Such characteristics are pivotal in understanding electric potentials associated with conductors and help ensure error-free calculations.