Problem 28 - 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) \(48.0 \mathrm{cm} ;\) (b) \(24.0 \mathrm{cm} ;(\mathrm{c}) 12.0 \mathrm{cm}.\)

Step-by-Step Solution

Answer

48 cm: 65.6 V; 24 cm: 131 V; at 12 cm: 131 V (same as at 24 cm).

Step 1: Understanding the Problem

We are tasked with finding the electric potential at various distances from the center of a charged metal sphere. The charge of the sphere is given (3.50 nC), and its radius is 24.0 cm. The metal sphere can be treated as a point charge at distances greater than or equal to its radius. Beyond this distance, the potential is given by \( V = \frac{kQ}{r} \), where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \mathrm{Nm^2/C^2} \), \( Q \) is the charge, and \( r \) is the distance from the center of the sphere.

Step 2: Calculate Potential at 48.0 cm

Since 48.0 cm is greater than the sphere's radius, we use the formula \( V = \frac{kQ}{r} \) to calculate potential. Convert the charge \( Q = 3.50 \mathrm{nC} = 3.50 \times 10^{-9} \mathrm{C} \) and the distance \( r = 48.0 \mathrm{cm} = 0.48 \mathrm{m} \). Thus:\[V = \frac{(8.99 \times 10^9 \mathrm{Nm^2/C^2})(3.50 \times 10^{-9} \mathrm{C})}{0.48 \mathrm{m}}\]Calculating this gives the potential at 48.0 cm.

Step 3: Calculate Potential at 24.0 cm

At 24.0 cm, the distance is equal to the sphere's radius, so the entire charge appears as though it is concentrated at the center. We again use \( V = \frac{kQ}{r} \), setting \( r = 0.24 \mathrm{m} \):\[V = \frac{(8.99 \times 10^9 \mathrm{Nm^2/C^2})(3.50 \times 10^{-9} \mathrm{C})}{0.24 \mathrm{m}}\]Calculating this gives the potential at 24.0 cm.

Step 4: Calculate Potential at 12.0 cm

Inside the conducting sphere (where 12.0 cm < 24.0 cm), the electric potential remains constant at the value it has at the surface of the sphere. Thus, the potential at 12.0 cm is the same as at 24.0 cm. Therefore, the potential at 12.0 cm is given by the same value found in Step 3.

Key Concepts

Coulomb's Law

Coulomb's Law plays a crucial role in understanding the interactions between electric charges. It states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:\[F = k \frac{|q_1 q_2|}{r^2}\]where:

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

This law forms the basis for calculating electric forces in electrostatics and is crucial in determining the electric potential around charged objects like spheres.

Electric Charge Distribution

Electric charge distribution refers to how electric charges are spread across a surface. In the given problem, we have a uniform distribution of charge across a metal sphere. This means every point on the sphere's surface has the same amount of charge.A uniformly charged metal sphere can be treated as if all the charge is concentrated at its center once you are outside the sphere. This simplification allows us to use the formula for electric potential outside the sphere:\[V = \frac{kQ}{r}\]where:

• \( V \) is the electric potential.
• \( k \) is Coulomb's constant.
• \( Q \) is the total charge of the sphere.
• \( r \) is the distance from the center of the sphere where potential is being measured.

Conductors

Conductors are materials that allow electric charges to flow freely. Metals are typical examples, with electrons moving easily within them. In a conductor like the metal sphere in our problem, the charges distribute evenly on the surface. Inside a conductor in electrostatic equilibrium, the electric field is zero. This means any point inside the conductor, closer to its surface, will have the same electric potential due to the redistribution of surface charges. This explains why the potential inside the sphere remains the same as at its surface.

Potential Energy

Potential energy in electrostatics is related to the position of a charge within an electric field. It represents the work done to move a charge from one point to another within this field.The electric potential (often simply called potential) is defined as the potential energy per unit charge. It provides insight into how a charged object will behave within the field. For a charge \( Q \) at distance \( r \) from a sphere, the potential energy formula is related to the potential:\[U = qV\]where \( U \) is the potential energy, \( q \) is the test charge, and \( V \) is the electric potential.

Electrostatics

Electrostatics is the study of electric charges at rest. It covers the forces, fields, and potentials associated with stationary charges. In electrostatics, we primarily use concepts like Coulomb's Law and electric charge distribution. In the provided exercise, electrostatics principles help determine the potential around a charged sphere. It's essential to consider how charges distribute on conductors and how they affect the electric field and potential. Understanding these basics enables calculations and predictions about electric interactions around objects.