Problem 45
Question
22.45. Concentric Spherical Shells. A small conducting spherical shell with inner radius \(a\) and outer radius \(b\) is concentric with a larger conducting spherical shell with inner radius \(c\) and outer radius \(d\) (Fig. 22.39\()\) . The inner shell has total charge \(+2 q,\) and the outer shell has charge \(+4 q\) . (a) Calculate the electric field (magnitude and direction) in terms of \(q\) and the distance \(r\) from the common center of the two shells for (i) \(rd\) . Show your results in a graph of the radial component of \(\vec{E}\) as a function of \(r .\) (b) What is the total charge on the (i) inner surface of the small shell; (ii) outer surface of the small shell; (iii) inner sur-=face of the large shell; (iv) outer surface of the large shell?
Step-by-Step Solution
Verified Answer
Electric fields are zero inside conductors, and vary outward as \(\frac{q}{\varepsilon_0 r^2}\). Total charges are \(0, +2q, -2q, +6q\) for the specified surfaces.
1Step 1: Electric Field for r < a
For the region inside the inner shell where \( r < a \), since the space is completely enclosed by a conductor, by Gauss's law, the electric field must be zero. Inside a conductor, the electric field is always zero because charges reside on the surface.
2Step 2: Electric Field for a < r < b
For the region \( a < r < b \), we consider a Gaussian surface within the thickness of the inner shell. Inside a conductor, the electric field is zero due to charge redistribution.
3Step 3: Electric Field for b < r < c
In the region \( b < r < c \), the electric field is due to the charge \(+2q\) from the inner shell. Using Gauss's law, by drawing a Gaussian sphere of radius \(r\), the electric field \(E\) is given by:\[E \cdot 4\pi r^2 = \frac{2q}{\varepsilon_0} \rightarrow E = \frac{2q}{4\pi \varepsilon_0 r^2}\]
4Step 4: Electric Field for c < r < d
For the region \(c < r < d\), the net charge enclosed by a Gaussian surface is both \(+2q\) from the small sphere and \(+4q\) from the inner surface of the larger shell. Hence, the total charge is \(+2q + 4q = +6q\). So, the electric field \(E\) is:\[E \cdot 4\pi r^2 = \frac{6q}{\varepsilon_0} \rightarrow E = \frac{6q}{4\pi \varepsilon_0 r^2}\]
5Step 5: Electric Field for r > d
Beyond the outer shell \( r > d \), the charges are enclosed within our Gaussian surface. Thus, the only contribution to the electric field comes from the total charge of the system \(+6q\). Therefore, the electric field \(E\) is:\[E \cdot 4\pi r^2 = \frac{6q}{\varepsilon_0} \rightarrow E = \frac{6q}{4\pi \varepsilon_0 r^2}\]
6Step 6: Charge on Inner Surface of Small Shell
The charge on the inner surface of the small shell will adjust to nullify any internal electric field, which means no net charge is present inside a conductor's cavity. Therefore, it is \(0\).
7Step 7: Charge on Outer Surface of Small Shell
The outer surface must contain the total charge of the small shell \(+2q\), as there is no internal charge within the conducting material to offset this total.
8Step 8: Charge on Inner Surface of Large Shell
The inner surface of the large shell must exactly cancel the electric field in the region between the shells \(b < r < c\), resulting in a charge of \(-2q\) to cancel the \(+2q\) from the small shell.
9Step 9: Charge on Outer Surface of Large Shell
The outer surface of the large shell will have a charge combining the original charge \(+4q\) with the inner surface's \(-2q\), resulting in a net charge of \(+6q\) on the outer surface.
Key Concepts
Gaussian surfaceConducting spherical shellElectric field
Gaussian surface
In electrostatics, a Gaussian surface is an imaginary boundary used to simplify the analysis of electric fields. This concept is especially useful when applying Gauss's law, which links electric fields to the charge enclosed within a surface. Imagine drawing an imaginary sphere around the charge distribution; this sphere is your Gaussian surface. It helps us determine how much "electric field" leaks out from the charge.
To use a Gaussian surface effectively, we need to ensure symmetry. In our exercise, the use of a Gaussian sphere is ideal because it perfectly matches the symmetry of spherical shells.
Although the process might sound abstract, it's a powerful tool that often dramatically simplifies finding electric fields, something you'll see demonstrated with the concentric shells exercise.
To use a Gaussian surface effectively, we need to ensure symmetry. In our exercise, the use of a Gaussian sphere is ideal because it perfectly matches the symmetry of spherical shells.
Although the process might sound abstract, it's a powerful tool that often dramatically simplifies finding electric fields, something you'll see demonstrated with the concentric shells exercise.
Conducting spherical shell
A conducting spherical shell is like a hollow ball made of electrically conductive material. The significant feature of such shells is that they exhibit some unique electrical behaviors due to their material properties.
Firstly, in conductors, charges distribute themselves on the outer surfaces, leaving the electric field inside the conductor zero. This is why, in our exercise, the electric field is zero for regions inside the conducting shells themselves.
In the specific case of concentric spherical shells, the behavior of charges becomes even more interesting as charges on one shell can influence charge distribution on the other, even if they're not physically in contact, through electric fields and induction effects.
The exercise demonstrates how each conducting shell affects the electric field in different regions based on charge distributions.
Firstly, in conductors, charges distribute themselves on the outer surfaces, leaving the electric field inside the conductor zero. This is why, in our exercise, the electric field is zero for regions inside the conducting shells themselves.
In the specific case of concentric spherical shells, the behavior of charges becomes even more interesting as charges on one shell can influence charge distribution on the other, even if they're not physically in contact, through electric fields and induction effects.
The exercise demonstrates how each conducting shell affects the electric field in different regions based on charge distributions.
Electric field
The electric field is a region around a charged particle or object within which a force would be exerted on other charges. Think of it as the "influence" a charge exerts on its surrounding area.
In the context of concentric spherical shells, the electric field varies depending on the region being considered. By using Gauss’s law, we can calculate the electric field in different sections outside and inside the shells.
For instance, within the thickness of the shells, because it’s conducting, the electric field remains zero. For regions like between the shells or beyond them, the electric field can be calculated using the formula:\[E = \frac{Q}{4\pi \varepsilon_0 r^2}\]Here, \(Q\) is the enclosed charge, \(\varepsilon_0\) is the permittivity of free space, and \(r\) is the radial distance from the center.
Understanding these regions is key to mastering electrostatics and predicting how charged bodies interact!
In the context of concentric spherical shells, the electric field varies depending on the region being considered. By using Gauss’s law, we can calculate the electric field in different sections outside and inside the shells.
For instance, within the thickness of the shells, because it’s conducting, the electric field remains zero. For regions like between the shells or beyond them, the electric field can be calculated using the formula:\[E = \frac{Q}{4\pi \varepsilon_0 r^2}\]Here, \(Q\) is the enclosed charge, \(\varepsilon_0\) is the permittivity of free space, and \(r\) is the radial distance from the center.
Understanding these regions is key to mastering electrostatics and predicting how charged bodies interact!
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