Chapter 22: Problem 14
A solid metal sphere with radius 0.450 m carries a net charge of 0.250 nC. Find the magnitude of the electric field (a) at a point 0.100 m outside the surface of the sphere and (b) at a point inside the sphere, 0.100 m below the surface.
Short Answer
Expert verified
(a) 7.39 N/C outside; (b) 0 N/C inside.
Step by step solution
01
Understanding the Problem
We have a metal sphere with a radius of 0.450 m carrying a net charge of 0.250 nC. We need to find the electric field at two different points. Firstly, at 0.100 m outside the sphere's surface and secondly, at 0.100 m inside the sphere's surface.
02
Using Gauss's Law for Outside Field Calculation
For a point outside a charged sphere, the sphere can be considered a point charge located at its center. The electric field at any distance, r, from the center is determined using Gauss's law:\[ E = \frac{kQ}{r^2} \]where \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\), \(Q\) is the net charge, and \(r\) is the distance from the center of the sphere to the point, which is the sphere's radius plus the distance outside the surface.
03
Calculate Total Distance from the Center for Outside Point
The radius of the sphere is 0.450 m, and the point is 0.100 m outside the surface, so the total distance from the center is:\[ r = 0.450 + 0.100 = 0.550 \, \text{m} \].
04
Calculate Electric Field at 0.550 m Outside
Plug in the values into the formula for electric field:\[ E = \frac{(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2)(0.250 \times 10^{-9} \, \text{C})}{(0.550 \, \text{m})^2}\]Calculating this gives:\[ E \approx 7.39 \, \text{N/C} \].
05
Understanding the Electric Field Inside a Conductor
For the point inside the conducting sphere, the electric field is zero because the electric charges redistribute themselves across the surface of the sphere, canceling the field inside in a static situation.
06
Conclusion
For a point outside the sphere (0.550 m from the center), the electric field magnitude is \(7.39\, \text{N/C}\). For a point inside the sphere, the electric field is \(0 \, \text{N/C}\) by the property of conductors at electrostatic equilibrium.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gauss's Law
Gauss's Law is a powerful principle used in electromagnetism to relate the electric field distribution to the charge distribution within a closed surface. According to Gauss's Law, the total electric flux through a closed surface is proportional to the enclosed charge. Mathematically, this can be expressed as:
\[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \]where \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is a vector representing an infinitesimal area on the closed surface, \( Q_{enc} \) is the total charge enclosed within the surface, and \( \varepsilon_0 \) is the permittivity of free space.
This concept is particularly useful for symmetrical charge distributions, like a charged sphere, as it simplifies calculations significantly by allowing us to treat the sphere as a point charge, especially when calculating the field outside the sphere.
\[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \]where \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is a vector representing an infinitesimal area on the closed surface, \( Q_{enc} \) is the total charge enclosed within the surface, and \( \varepsilon_0 \) is the permittivity of free space.
This concept is particularly useful for symmetrical charge distributions, like a charged sphere, as it simplifies calculations significantly by allowing us to treat the sphere as a point charge, especially when calculating the field outside the sphere.
- Only the charge enclosed by the Gaussian surface affects the electric field.
- The symmetry of the charge distribution simplifies the application of Gauss's Law.
metal sphere charge distribution
In the context of a metal sphere, charge distribution is an interesting topic because of the manner charges spread across conductive materials. When a metal sphere is charged, the charges (typically electrons) move freely until equilibrium is reached. In electrostatic equilibrium:
The charges being on the surface is why the calculation for electric field inside the sphere typically yields zero, as stated in the step-by-step solution. Since there are no charges inside to contribute to a field, the net field within remains zero.
- All excess charges reside on the surface of the conductor.
- The electric field inside a conductor is zero when in a steady state.
The charges being on the surface is why the calculation for electric field inside the sphere typically yields zero, as stated in the step-by-step solution. Since there are no charges inside to contribute to a field, the net field within remains zero.
Coulomb's constant
Coulomb's constant, denoted as \( k \), is a value that arises from Coulomb's Law, which describes the electrostatic interaction between electric charges. It is defined as:
\[ k = 8.99 \times 10^{9} \, \text{Nm}^2/\text{C}^2 \]Coulomb's Law uses this constant to describe the force between two charges, and it also applies when calculating electric fields around charged objects:
\[ F = k \frac{|q_1 q_2|}{r^2} \]where \( F \) is the force between two point charges \( q_1 \) and \( q_2 \), and \( r \) is the distance between them.
This constant is crucial in calculating the electric field using Gauss's Law outside the sphere. By treating the charged sphere as a point charge at its center, we effectively apply Coulomb's constant to find the electric field at any point outside the sphere, as per the problem solution. This is possible because of the symmetry of the sphere.
\[ k = 8.99 \times 10^{9} \, \text{Nm}^2/\text{C}^2 \]Coulomb's Law uses this constant to describe the force between two charges, and it also applies when calculating electric fields around charged objects:
\[ F = k \frac{|q_1 q_2|}{r^2} \]where \( F \) is the force between two point charges \( q_1 \) and \( q_2 \), and \( r \) is the distance between them.
This constant is crucial in calculating the electric field using Gauss's Law outside the sphere. By treating the charged sphere as a point charge at its center, we effectively apply Coulomb's constant to find the electric field at any point outside the sphere, as per the problem solution. This is possible because of the symmetry of the sphere.