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Magazine Chapter 22: Problem 18
(II) A solid metal sphere of radius \(3.00 \mathrm{~m}\) carries a total charge of \(-5.50 \mu \mathrm{C}\). What is the magnitude of the electric field at a distance from the sphere's center of (a) \(0.250 \mathrm{~m}\), (b) \(2.90 \mathrm{~m},\) (c) \(3.10 \mathrm{~m},\) and \((d) 8.00 \mathrm{~m} ?\) How would the answers differ if the sphere were \((e)\) a thin shell, or \((f)\) a solid nonconductor uniformly charged throughout?
Short Answer
Expert verified
(a) 0 N/C, (b) 0 N/C, (c) 5.13 x 10^4 N/C, (d) 7.73 x 10^3 N/C; (e) same as (c) and (d) for r > 3m, (f) same for r > 3m.
Step by step solution
01
Understanding the Problem
We need to find the electric field at various distances from the center of a charged sphere with a specified charge and radius. Gauss's Law will help us determine the electric field at a point based on the symmetry of the problem.
02
Applying Gauss's Law
Gauss's Law states that the electric flux through a closed surface is proportional to the charge enclosed, given by \( \Phi_E = \frac{Q_{enc}}{\varepsilon_0} \). For a sphere, the electric field \( E \) at any point outside or inside the sphere can be determined by symmetry, using the formula \( E \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0} \), where \( Q \) is the total enclosed charge and \( r \) is the radial distance from the center.
03
Case (a): Inside the Sphere (r = 0.250 m)
At \( r = 0.250 \) m, which is inside the metal sphere, the electric field is zero. This follows because the charges redistribute on the surface of a conductor, leading to no electric field inside a charged conductor.
04
Case (b): Inside the Sphere (r = 2.90 m)
Like in case (a), since \( r = 2.90 \) m is still less than the sphere's radius (3.00 m), the electric field is still zero because the point is inside the conductive sphere.
05
Case (c): On the Surface (r = 3.10 m)
Now \( r = 3.10 \) m is outside the sphere, thus we use the formula \( E = \frac{Q}{4\pi\varepsilon_0 r^2} \). Substitute \( Q = -5.50 \times 10^{-6} \) C, \( r = 3.10 \) m, and \( \varepsilon_0 = 8.85 \times 10^{-12} \) C虏/(N路m虏) to find the electric field: \[ E = \frac{-5.50 \times 10^{-6}}{4\pi(8.85 \times 10^{-12})(3.10)^2} \approx 5.13 \times 10^4 \text{ N/C} \]
06
Case (d): Far from the Sphere (r = 8.00 m)
At this distance, \( r = 8.00 \) m, we again use the formula for points outside the sphere: \( E = \frac{Q}{4\pi\varepsilon_0 r^2} \). Substitute \( r = 8.00 \) m: \[ E = \frac{-5.50 \times 10^{-6}}{4\pi(8.85 \times 10^{-12})(8.00)^2} \approx 7.73 \times 10^3 \text{ N/C} \]
07
Case (e): If the Sphere is a Thin Shell
For a thin shell, the electric field is zero inside the shell (\( r < 3 \text{ m} \)). For \( r > 3 \text{ m} \), the electric field at points in space depends only on the total charge and is calculated the same way as we did for \( r = 3.10 \text{ m} \) and \( r = 8.00 \text{ m} \).
08
Case (f): Solid Nonconductor
If the sphere is a non-conducting solid, the electric field is non-zero inside, calculated differently. However, outside (\( r > 3 \text{ m} \)), the field is the same as for a conductor or shell, following \( E = \frac{Q}{4\pi\varepsilon_0 r^2} \).
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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 fundamental principle in electromagnetism that relates the electric flux through a closed surface to the charge enclosed within that surface. Simply put, it allows you to determine the electric field around symmetrical charge distributions. This law is especially useful in situations where the symmetry of the charge distribution simplifies the calculations.
- Mathematically, Gauss's Law is expressed as \( \Phi_E = \frac{Q_{enc}}{\varepsilon_0} \), where \( \Phi_E \) is the electric flux, \( Q_{enc} \) is the enclosed charge, and \( \varepsilon_0 \) is the vacuum permittivity.
- For a spherical charge distribution, we rely on symmetry to state that the electric field is uniform over a spherical surface centered on the charge. This reduces the problem of calculating the electric field to a simple expression.
- By enclosing the charge in a hypothetical spherical surface (a Gaussian surface), it's easier to find the electric field at any point around the charge using the formula \( E \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0} \).
Conductors and Insulators
Conductors and insulators are materials that respond differently to electric charge. Understanding their behavior is crucial when examining electric fields, especially within and around spherical objects.
- Conductors are materials that allow charges to flow freely. In a conductor, electric charges redistribute until the electric field within it is zero, leading to an equilibrium. This is why the electric field inside a conductive sphere is zero, as seen in examples when determining electric fields at points inside a metal sphere.
- Insulators, also known as dielectrics, do not allow for free charge movement. The behavior of electric fields inside or outside an insulating sphere differs, since charge doesn't redistribute easily. In a non-conductive solid sphere, the electric field inside follows a different pattern because of the uniform charge distribution throughout the material.
Electric Flux
Electric flux is a measure of the number of electric field lines passing through a given surface. It is a crucial concept in understanding the flow of electric fields through space.
- Represented by \( \Phi_E \), electric flux depends on both the electric field strength and the area through which the field lines pass. In Gauss's Law, it's directly related to the enclosed charge.
- The integral form of calculating electric flux is \( \Phi_E = \int \mathbf{E} \cdot \mathbf{dA} \), where \( \mathbf{E} \) is the electric field vector and \( \mathbf{dA} \) is a differential area vector on the closed surface.
- In uniform fields and over symmetrical surfaces such as spheres, the calculation simplifies considerably. The symmetry often allows factorizing the integral into a simple multiplication of electric field and total area.
Charge Distribution
Charge distribution describes how electric charge is concentrated or spread out over a region or surface. Understanding charge distribution is key to analyzing the electric field produced by various objects.
- On conductive surfaces, charges redistribute evenly across surfaces to minimize repulsion. For this reason, in a charged conductive sphere, the charge resides on the outer surface, creating no electric field inside.
- In insulative materials, charge may be distributed uniformly throughout the volume. This alters how the electric field behaves both inside and outside the sphere.
- The type of charge distribution (surface or volume) influences how we apply Gauss's Law, especially for determining electric fields at different points relative to the charged object.
