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Two isolated, concentric, conducting spherical shells have radii \(R_{1}=0.500 \mathrm{~m}\) and \(R_{2}=1.00 \mathrm{~m}\), uniform charges \(q_{1}=+2.00\) \(\mu \mathrm{C}\) and \(q_{2}=+1.00 \mu \mathrm{C}\), and negligible thicknesses. What is the magnitude of the electric field \(E\) at radial distance (a) \(r=4.00 \mathrm{~m}\), (b) \(r=0.700 \mathrm{~m}\), and \((\mathrm{c}) r=0.200 \mathrm{~m} ?\) With \(V=0\) at infinity, what is \(V\) at (d) \(r=4.00 \mathrm{~m},(\mathrm{e}) r=1.00 \mathrm{~m},(\mathrm{f}) r=0.700 \mathrm{~m},(\mathrm{~g}) r=0.500 \mathrm{~m}\) (i) \(r=0 ?\) (h) \(r=0.200 \mathrm{~m}\), and (j) Sketch \(E(r)\) and \(V(r)\).

Short Answer

Expert verified
The electric field is zero at and inside the inner shell, decreasing outside. Voltage decreases as \( r \) increases. Electric field graphs show steps; voltage graphs slope down with radius.

Step by step solution

01

Understanding Gauss's Law

Gauss's Law states that the electric field \( E \) times the surface area of a closed surface is equal to the charge enclosed \( Q_{enclosed} \) divided by the permittivity of free space \((\varepsilon_0)\), i.e., \(abla \cdot \mathbf{E} = rac{Q_{enclosed}}{\varepsilon_0}.\) This will be used to determine the electric field at different radial distances.
02

Electric Field at r = 4.00 m

At \( r = 4.00 \ m\), the Gaussian surface encloses both conducting charges \( q_1 \) and \( q_2 \). Thus, the total enclosed charge \( Q_{enclosed} = q_1 + q_2 = 2.00 \ \mu C + 1.00 \ \mu C = 3.00 \ \mu C\). Using Gauss's Law, \( E = rac{Q_{enclosed}}{4\pi \varepsilon_0 r^2} = rac{3.00 \times 10^{-6} \ C}{4 \pi \left(8.854 \times 10^{-12} \ C^2/N \cdot m^2\right)(4.00 \ m)^2} \approx 1.69 \times 10^{3} \ N/C\).
03

Electric Field at r = 0.700 m

At \( r = 0.700 \ m\), the Gaussian surface encloses only the inner charge \( q_1 = 2.00 \ \mu C\). By Gauss's Law, the electric field \( E = rac{q_1}{4 \pi \varepsilon_0 r^2} = rac{2.00 \times 10^{-6} \ C}{4 \pi \left(8.854 \times 10^{-12} \ C^2/N \cdot m^2\right)(0.700 \ m)^2} \approx 3.67 \times 10^{4} \ N/C\).
04

Electric Field at r = 0.200 m

At \( r = 0.200 \ m\), the point is inside the inner shell, which is a conductor. Hence, the electric field inside a conductor is zero, \( E = 0\).
05

Voltage at r = 4.00 m

Voltage at a point \( r \) can be calculated by integrating the electric field from \( r \) to infinity. Since \( V(\infty) = 0\), we have \( V(4.00 \ m) = rac{3.00 \times 10^{-6} \ C}{4 \pi \varepsilon_0 \cdot 4.00} = 6.75 \times 10^{3} \ V\).
06

Voltage at r = 1.00 m

For \( r = 1.00 \ m\), the potential is influenced by both the enclosed charges. By integration from \( R_2 \) to \( \infty\), and using \( q_1+q_2 = 3 \ \mu C\), the voltage \( V(1.00 \ m) = 2.70 \times 10^{4} \ V\).
07

Voltage at r = 0.700 m

For \( r = 0.700 \ m\), the potential has contributions from both shells, as it is less than the outer shell radius. Using similar integration procedures, \( V(0.700 \ m) = 5.00 \times 10^{4} \ V\).
08

Voltage at r = 0.500 m

At the inner surface \( r = 0.500 \ m\), the point is at the boundary of the inner shell, where potential is constant inside a conductor: \( V(0.500 \ m) = 5.13 \times 10^{4} \ V\).
09

Voltage at r = 0

At \( r = 0\), still inside the conductor, the electric potential is constant and is the same as that at the surface of the inner shell: \( V(0) = 5.13 \times 10^{4} \ V\).
10

Voltage at r = 0.200 m

Again, since \( r = 0.200 \ m \) is within the inner conductor, the electric potential remains constant, \( V(0.200 \ m) = 5.13 \times 10^{4} \ V\).
11

Plotting E(r) and V(r)

For \( E(r)\), plot step-like behavior: 0 inside inner shell, \( E \propto \frac{1}{r^2}\) between shells, and \( E \propto \frac{1}{r^2}\) beyond outer shell. For \( V(r)\), plot decreasing \( \propto \frac{1}{r}\) outside shells, constant inside shells.

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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 key principle for understanding electric fields around symmetrical charge distributions. It relates the electric field flowing out of a closed surface to the charge enclosed within the surface. If you imagine a closed shape, like a sphere, any charge inside will contribute to the electric field flowing through its surface. Gauss's Law is mathematically expressed as: \[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \]Here, \( \Phi_E \) is the electric flux, \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is a differential area on the closed surface, and \( Q_{\text{enclosed}} \) is the total charge within that surface. Permittivity of free space, \( \varepsilon_0 \), is a constant that describes how electric fields interact with the vacuum. When dealing with concentric spherical shells, Gauss's Law simplifies calculations because the electric field at a certain distance from the center will only depend on the charge within that radius. For the distances given in the exercise, consider the electric field produced by the enclosed charges as you apply Gauss's Law.
Voltage in Conductors
Voltage, or electric potential, in conductors plays a crucial role in understanding how charges behave in such materials. A conductor is a material that allows free movement of electric charges within its surface. In electrostatic equilibrium, charges reside on the surface, implying that the electric field inside a conductor is zero. Therefore, the voltage, or electric potential, throughout the conductor is constant. This constant voltage in a conductor is because work is needed to move a charge against an electric field. In a conductor with no internal electric field, no work is required to move a charge, resulting in a uniform potential. This principle explains why, in a system of concentric spherical shells, the potential remains the same inside the inner shell and between the shells until you reach the outer surface. Furthermore, the voltage can be measured relative to a reference point, often set at infinity, simplifying the integration involved when calculating potential differences.
Electric Potential
Electric potential, often referred to as voltage, is the work done per unit charge in bringing a test charge from a reference point to a specific point in an electric field without accelerating it. It's a scalar quantity that tells us about the energy landscape created by electric charges. The formula for calculating potential difference, \( V \), between two points is given by:\[ V = - \int_{r_1}^{r_2} \mathbf{E} \cdot d\mathbf{r} \]In scenarios with spherical symmetry, such as with concentric spherical shells, the calculation can be further simplified by directly integrating the electric field along the radial direction. The influence of both charges needs to be accounted for potential calculations.When solving problems like the one in this exercise, you're usually given a reference point where potential is zero, such as at infinity. Any potential at a point is then calculated relative to that zero point, allowing for easy calculation and comparison of potential values.
Concentric Spherical Shells
Concentric spherical shells offer an excellent model for learning about electric fields and potentials in symmetrical arrangements. They consist of multiple spherical layers sharing the same center point, like nested balls. By applying Gauss's Law and considering symmetrical charge distributions, the electric field and potential can be determined for any point inside or outside these shells.
  • Inside the inner shell: Here, the electric field is zero due to the shell being a conductor. Applying Gauss's Law confirms that no charge is enclosed, thus no field.
  • Between the shells: In this region, only the charge on the inner shell affects the electric field, acting as though all charge were concentrated at the center. The field follows the inverse square law, decreasing with distance.
  • Outside the outer shell: Here, the enclosed charge is the sum of charges on both shells. Again, the field behaves as though the entire charge is concentrated at the center.
The potential difference at any point can also be calculated by integrating the electric field. The spherical symmetry allows for straightforward computational applications using the assumption that the charges are point charges situated at the center.

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Most popular questions from this chapter

An electron is projected with an initial speed of \(3.2 \times 10^{5} \mathrm{~m} / \mathrm{s}\) directly toward a proton that is fixed in place. If the electron is initially a great distance from the proton, at what distance from the proton is the speed of the electron instantaneously equal to twice the initial value?

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