Problem 60
Question
Two spherical shells have a common center. The inner shell has radius \(R_1 =\) 5.00 cm and charge \(q1 = +3.00 \times 10^{-6}\) C; the outer shell has radius \(R_2 =\) 15.0 cm and charge \(q2 = -5.00 \times 10^{-6}\) C. Both charges are spread uniformly over the shell surface. What is the electric potential due to the two shells at the following distances from their common center: (a) \(r =\) 2.50 cm; (b) \(r =\) 10.0 cm; (c) \(r =\) 20.0 cm? Take \(V = 0\) at a large distance from the shells.
Step-by-Step Solution
Verified Answer
V = 539400 V at r = 2.50 cm, V = -29966.67 V at r = 10.0 cm, V = -89900 V at r = 20.0 cm.
1Step 1: Understanding the Problem
We have two spherical shells with a common center. We need to find the electric potential due to these shells at different distances from their common center. The electric potential is influenced by the given radii and charges of the shells.
2Step 2: Electric Potential Formula
The electric potential due to a point charge is given by the formula \( V = \frac{k \, q}{r} \), where \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge.
3Step 3: Potential at r = 2.50 cm
At \( r = 2.50 \text{ cm} \), we are inside the inner shell. The potential due to the inner shell is constant and is \( V_1 = \frac{k \cdot q1}{R_1} \). The potential due to the outer shell is negligible because it's uniform over the shell, \( V_2 = 0 \). Thus, \( V = V_1 + V_2 = \frac{k \cdot 3.00 \times 10^{-6}}{0.05} \).
4Step 4: Calculate Potential at r = 2.50 cm
Substitute the values in the formula: \[ V = \frac{8.99 \times 10^9 \cdot 3.00 \times 10^{-6}}{0.05} = 539400 \, \text{V} \].
5Step 5: Potential at r = 10.0 cm
At \( r = 10.0 \text{ cm} \), between the two shells, the potential from the inner shell is \( V_1 = \frac{k \cdot 3.00 \times 10^{-6}}{0.10} \) and from the outer shell is \( V_2 = \frac{k \cdot -5.00 \times 10^{-6}}{0.15} \). Thus, \( V = V_1 + V_2 \).
6Step 6: Calculate Potential at r = 10.0 cm
Substitute the values in the formula for each shell: \[ V_1 = \frac{8.99 \times 10^9 \cdot 3.00 \times 10^{-6}}{0.10} = 269700 \, \text{V} \]\[ V_2 = \frac{8.99 \times 10^9 \cdot -5.00 \times 10^{-6}}{0.15} = -299666.67 \, \text{V}\] Therefore, \( V = 269700 - 299666.67 = -29966.67 \, \text{V} \).
7Step 7: Potential at r = 20.0 cm
At \( r = 20.0 \text{ cm} \), outside both shells. Use the total charge \( q_{total} = q1 + q2 = -2.00 \times 10^{-6} \text{C} \). The potential is given by \( V = \frac{k \cdot q_{total}}{r} \).
8Step 8: Calculate Potential at r = 20.0 cm
Substitute the values in the formula: \[ V = \frac{8.99 \times 10^9 \cdot -2.00 \times 10^{-6}}{0.20} = -89900 \, \text{V} \].
Key Concepts
Spherical ShellsCoulomb's LawPoint ChargeElectric Field
Spherical Shells
In physics, spherical shells are important concepts when analyzing electric fields and potentials involving symmetrical objects. A spherical shell is essentially a hollow sphere, with the charges distributed uniformly on its surface. Understanding the behavior of electric fields and potentials inside and outside these shells is crucial.
One key characteristic of spherical shells is that within a uniformly charged shell, the electric field is zero. This occurs because the charges are evenly distributed, causing their field effects inside the shell to cancel each other. As a result, when considering problems involving the electric potential of spherical shells, the potential inside the shell remains constant.
Outside the shell, the situation changes. Here, the electric field behaves as though all the charge is concentrated at the center. This simplifies calculations because the problem can be treated as a point charge, allowing us to use equations derived from Coulomb's Law.
One key characteristic of spherical shells is that within a uniformly charged shell, the electric field is zero. This occurs because the charges are evenly distributed, causing their field effects inside the shell to cancel each other. As a result, when considering problems involving the electric potential of spherical shells, the potential inside the shell remains constant.
Outside the shell, the situation changes. Here, the electric field behaves as though all the charge is concentrated at the center. This simplifies calculations because the problem can be treated as a point charge, allowing us to use equations derived from Coulomb's Law.
Coulomb's Law
Coulomb's Law is a fundamental principle that describes the force between two point charges. It states that the electric force (\( F \)) between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. The formula for Coulomb's Law is:
\( F = k \frac{|q_1 \, q_2|}{r^2} \)
where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \). This law not only helps calculate forces but also electric potentials and fields.
When adapting Coulomb's Law for potential calculations, the equation changes slightly to focus on potential energy instead of force. The potential (\( V \)) due to a point charge is given by:
\( V = k \frac{q}{r} \)
This equation is especially useful when analyzing problems with spherical shells, as each shell's potential can be calculated using its respective charge and radius.
\( F = k \frac{|q_1 \, q_2|}{r^2} \)
where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \). This law not only helps calculate forces but also electric potentials and fields.
When adapting Coulomb's Law for potential calculations, the equation changes slightly to focus on potential energy instead of force. The potential (\( V \)) due to a point charge is given by:
\( V = k \frac{q}{r} \)
This equation is especially useful when analyzing problems with spherical shells, as each shell's potential can be calculated using its respective charge and radius.
Point Charge
A point charge is a term used in physics to simplify the concept of electric charge. It refers to a charge that is concentrated at a single point in space, with no spatial extent. In reality, all charges have some size, but many problems assume point charges for simplicity.
Point charges are fundamental because they allow us to apply Coulomb's Law directly for calculations involving forces, fields, or potentials. In the context of the spherical shell exercise, both shells can be viewed as having point charges at their centers when calculating potentials at points outside the shells.
Understanding point charges is crucial for visualizing scenarios in electrostatics, as they provide a clear, simplified model for analyzing interactions between charges. By treating larger distributions of charge as a series of point charges, complex problems become easier to handle.
Point charges are fundamental because they allow us to apply Coulomb's Law directly for calculations involving forces, fields, or potentials. In the context of the spherical shell exercise, both shells can be viewed as having point charges at their centers when calculating potentials at points outside the shells.
Understanding point charges is crucial for visualizing scenarios in electrostatics, as they provide a clear, simplified model for analyzing interactions between charges. By treating larger distributions of charge as a series of point charges, complex problems become easier to handle.
Electric Field
An electric field is a region around a charged object where other charges experience a force. It is represented by the symbol \( \mathbf{E} \) and quantitatively defined as the force per unit charge:\[ \mathbf{E} = \frac{\mathbf{F}}{q} \]where \( \mathbf{F} \) is the electric force and \( q \) is the charge experiencing the force.
Understanding electric fields is essential for predicting how charges will interact. Inside a spherical shell, as earlier mentioned, the electric field is zero. However, outside the shell, the field behaves as if the entire charge of the shell were concentrated at its center.
This characteristic allows for straightforward calculations of electric potentials and fields when studying charged spherical shells. The knowledge that the field inside is zero and behaves as a point charge outside is essential for solving problems like those in the exercise, where electric potential at various points is determined by evaluating the field and charge distribution at those points.
Understanding electric fields is essential for predicting how charges will interact. Inside a spherical shell, as earlier mentioned, the electric field is zero. However, outside the shell, the field behaves as if the entire charge of the shell were concentrated at its center.
This characteristic allows for straightforward calculations of electric potentials and fields when studying charged spherical shells. The knowledge that the field inside is zero and behaves as a point charge outside is essential for solving problems like those in the exercise, where electric potential at various points is determined by evaluating the field and charge distribution at those points.
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