Problem 40 - University Physics with Modern Physics

Question

(a) How much excess charge must be placed on a copper sphere 25.0 cm in diameter so that the potential of its center, relative to infinity, is 3.75 kV? (b) What is the potential of the sphere 's surface relative to infinity?

Step-by-Step Solution

Answer

(a) Excess charge is approximately \( 5.21 \times 10^{-7} \text{ C} \). (b) Potential at the surface is 3.75 kV.

Step 1: Understand the Problem

We need to find the amount of excess charge required to achieve a specified electric potential of a sphere's center, and also find the potential at its surface. We'll use the formula for the electric potential due to a point charge, as a charged sphere behaves like a point charge when viewed from outside.

Step 2: Calculate the Radius of the Sphere

The sphere's diameter is given as 25.0 cm. We first convert this to meters to ensure our units are consistent. Then, we find the radius by dividing the diameter by 2. - Diameter = 25.0 cm = 0.25 m- Radius \( r = \frac{0.25}{2} = 0.125 \text{ m} \)

Step 3: Use the Formula for Electric Potential

The potential \( V \) at a distance \( r \) from a charge \( Q \) is given by: \[ V = \frac{kQ}{r} \]where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \). We substitute the given potential (= 3.75 kV = 3750 V) and radius, and solve for charge \( Q \).

Step 4: Solve for Excess Charge \( Q \)

Rearrange the formula to solve for \( Q \):\[ Q = \frac{Vr}{k} \]Substitute \( V = 3750 \text{ V} \), \( r = 0.125 \text{ m} \), and \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \):\[ Q = \frac{(3750)(0.125)}{8.99 \times 10^9} \approx 5.21 \times 10^{-7} \text{ C} \]

Step 5: Calculate the Potential at the Surface

For a sphere, the potential inside (at the center) is the same as at the surface. Therefore, the potential at the surface is also the same as the calculated center potential, which is 3.75 kV.

Key Concepts

Coulomb's Law

Coulomb's Law is fundamental to understanding the interactions between electric charges. It helps quantify the amount of force acting between charged particles. According to Coulomb's Law, the electric force (\( F \)) between two point charges (\( Q_1 \) and \( Q_2 \)) is directly proportional to the product of the quantities of charge and inversely proportional to the square of the distance (\( r \)) between them:

• The law is expressed mathematically as: \[ F = k \frac{|Q_1 Q_2|}{r^2} \]
• Where \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \) is the electrostatic constant, also known as Coulomb's constant.
• This principle is essential when calculating forces and potentials in electrostatic applications, such as in the exercise involving the charged sphere.

Unlike gravity, which only attracts, electric forces can attract or repel. This property makes Coulomb's Law incredibly useful for understanding a wide range of electrical phenomena. It provides a basis for understanding electric fields and potentials, especially relevant in scenarios involving charged spheres, like the one discussed here.

Excess Charge Calculation

Excess charge calculation is key when determining the total charge needed to create a particular electrical potential. In the context of our exercise, excess charge refers to the additional electrons that are either added or removed to create the desired electrical potential on a sphere.

• The formula used is \[ Q = \frac{Vr}{k} \]
• Where \( Q \) is the charge in coulombs, \( V \) is the potential in volts, \( r \) is the radius of the sphere in meters, and \( k \) is Coulomb's constant.
• This formula helps you determine how much charge must be added to or removed from a sphere in order to reach a specific electric potential of the sphere relative to a reference point, which typically is infinity in potential calculations.

In the exercise problem, we used the given potential and sphere radius to find the required excess charge. This process involves substituting the known values to reveal the charge needed to achieve the desired potential.

Charged Sphere Potential

The concept of charged sphere potential can be understood using the property that a charged sphere behaves like a point charge when viewed from the outside. This means that we can use similar calculations to find the electric potential of both.

• Potential of a charged sphere is uniform inside the sphere and is equal to the potential at its surface.
• This is due to the fact that any point inside a charged conductor in electrostatic equilibrium has the same potential as on its surface.
• This means once you determine the potential at the center of the sphere, you know the potential across the whole sphere and its surface.

The exercise demonstrates that the potential at the center of the charged sphere is the same as that at its outer surface. As calculated, both these potentials are the same as the given and desired potential value, 3.75 kV, showing that the potential of a uniformly charged conductor remains constant internally.