Question

A solid conducting sphere of radius \(r_{1}\) has a total charge of \(+3 Q .\) It is placed inside (and concentric with) a conducting spherical shell of inner radius \(r_{2}\) and outer radius \(r_{3}\). Find the electric field in these regions: \(rr_{3}\).

Short answer

Answer: The electric field in each of the four regions is: 1. For \(rr_{3}\): \(E(r>r_{3}) = \frac{3Q}{4\pi\varepsilon_{0}r^2}\)

Step-by-step solution

Understand Gauss's law

Gauss's law states that the electric flux through a closed surface is proportional to the charge enclosed by the surface. Mathematically, it can be written as: $$ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\varepsilon_{0}} $$ Here, \(\oint \vec{E} \cdot d\vec{A}\) is the electric flux through the closed surface, \(Q_{\text{enclosed}}\) is the charge enclosed by the surface, and \(\varepsilon_{0}\) is the electric constant.

Define Gaussian surfaces

For each region, we will define a Gaussian surface: a closed surface through which we will apply Gauss's law. We will use spherical surfaces because of the spherical symmetry of the problem. 1. For \(rr_{3}\), we will use a spherical surface with radius \(r\).

Apply Gauss's law to \(r

For \(r<r_{1}\), there is no enclosed charge because the Gaussian surface is inside the solid conducting sphere. According to Gauss's law, $$ \oint \vec{E} \cdot d\vec{A} = 0 $$ Since the electric field inside a conductor is always zero, we can conclude that the electric field in this region is: $$ E(r<r_{1})=0 $$

Apply Gauss's law to \(r_{1}

For \(r_{1}<r<r_{2}\), the Gaussian surface encloses only the solid conducting sphere with a charge of \(+3Q\). According to Gauss's law, $$ \oint \vec{E} \cdot d\vec{A} = \frac{3Q}{\varepsilon_{0}} $$ Since the electric field is radial and its magnitude is constant on a spherical surface, we can write the flux as \(E(r) \cdot 4\pi r^2 = \frac{3Q}{\varepsilon_{0}}\). Solving for \(E(r)\), we get $$ E(r_{1}<r<r_{2}) = \frac{3Q}{4\pi\varepsilon_{0}r^2} $$

Apply Gauss's law to \(r_{2}

For \(r_{2}<r<r_{3}\), the Gaussian surface is inside the conducting shell, which means there is no enclosed charge. According to Gauss's law, $$ \oint \vec{E} \cdot d\vec{A} = 0 $$ Since the electric field inside a conductor is always zero, we can conclude that the electric field in this region is: $$ E(r_{2}<r<r_{3})=0 $$

Apply Gauss's law to \(r>r_{3}\)

For \(r>r_{3}\), the Gaussian surface encloses the entire system, including both the solid sphere and the shell. Since the shell is neutral, it does not contribute any charge. So, the Gaussian surface only encloses the \(+3Q\) charge of the solid sphere. According to Gauss's law, $$ \oint \vec{E} \cdot d\vec{A} = \frac{3Q}{\varepsilon_{0}} $$ Since the electric field is radial and its magnitude is constant on a spherical surface, we can write the flux as \(E(r) \cdot 4\pi r^2 = \frac{3Q}{\varepsilon_{0}}\). Solving for \(E(r)\), we get $$ E(r>r_{3}) = \frac{3Q}{4\pi\varepsilon_{0}r^2} $$

Summarize the results

The electric field in each of the four regions can be summarized as follows: 1. For \(rr_{3}\): \(E(r>r_{3}) = \frac{3Q}{4\pi\varepsilon_{0}r^2}\)

Key concepts

Electric Charge

Electric charge is a fundamental property of matter that causes it to experience a force when near other electric charges. It comes in two types, which we refer to as positive and negative. Like charges repel each other while opposite charges attract. The unit of electric charge is the Coulomb (C).

The concept plays a crucial role when dealing with electrostatic problems, like the one in our exercise with a charged conducting sphere. Charges can be distributed on objects in various ways, and in conductors, they can move freely. In the problem given, the solid conducting sphere has a total charge of +3Q, which influences the electric field around it.

Conducting Sphere

A conducting sphere is an object that can distribute charge evenly across its surface due to the free movement of electrons. In electrostatic conditions, the interior of a conductor will have no excess charge - any excess charge resides on its surface. Therefore, the electric field inside a conductor at electrostatic equilibrium is zero. This is why, in our exercise, within the solid conducting sphere of radius r1, the electric field is zero (E(r
In a spherical shell, on the other hand, the charge distributes across the outer surface, affecting the electric field outside the sphere while keeping the inner region field-free as per Gauss's law.

Electric Flux

Electric flux is a measure of the amount of electric field passing through a given area. It's an essential concept in electromagnetism and is represented by the symbol \(\Phi_E\). The electric flux through an area is calculated as the electric field \(\vec{E}\) multiplied by the area of the surface \(\vec{A}\) and the cosine of the angle between the field and the normal to the surface. Mathematically, it's given by \(\Phi_E = \vec{E} \cdot \vec{A}\).

In cases with radial symmetry, such as around a spherical charge distribution, the angle is zero and the calculation simplifies to just the product of the electric field magnitude and the area.

Gaussian Surface

A Gaussian surface is an imaginary closed surface used in Gauss's law to calculate the electric field based on a symmetrical charge distribution. The choice of a Gaussian surface is crucial for simplifying problems in electrostatics. It should be chosen to exploit the symmetry of the setup, making it easier to evaluate the electric field.

For our spherical charge distributions, spherical Gaussian surfaces are chosen because the electric field at any point on such a surface is the same distance from the center, meaning the electric field has the same magnitude throughout, simplifying our calculations greatly.

Electric Constant

The electric constant, also known as the permittivity of free space (\(\varepsilon_0\)), is a physical constant that describes how electric fields interact with the vacuum of free space. Its value is approximately \(8.854 \times 10^{-12} \frac{C^2}{N \cdot m^2}\).

The electric constant is crucial in Gauss's law, representing a measure of how much the electric field can 'pass through' free space. In the solutions provided, it appears in the denominator of the equations for the electric field, indicating that the electric field strength is inversely proportional to the permittivity of free space, which relates the charge to the resulting electric field.