Question

A hollow conducting spherical shell has an inner radius of \(8.00 \mathrm{~cm}\) and an outer radius of \(10.0 \mathrm{~cm} .\) The electric field at the inner surface of the shell, \(E_{\mathrm{i}}\), has a magnitude of \(80.0 \mathrm{~N} / \mathrm{C}\) and points toward the center of the sphere, and the electric field at the outer surface, \(E_{\infty}\) has a magnitude of \(80.0 \mathrm{~N} / \mathrm{C}\) and points away from the center of the sphere (see the figure). Determine the magnitude of the charge on the inner surface and the outer surface of the spherical shell.

Short answer

Question: Determine the magnitude of the charge on the inner and outer surfaces of a hollow conducting spherical shell with inner and outer radii 8.00 cm and 10.0 cm, respectively. The magnitude of the electric field at the inner and outer surfaces is 80.0 N/C. Answer: The magnitude of the charge on the inner surface is approximately 2.26 x 10^(-11) C, while the magnitude of the charge on the outer surface is approximately 8.90 x 10^(-11) C.

Step-by-step solution

Formula for Electric Field of a Spherical Shell

We will use the formula for the electric field of a spherical shell given by: \(E = \dfrac{kQ}{r^2}\), Where E is the electric field, k is the electrostatic constant \((k = 8.99 × 10^9 \, Nm^2/C^2)\), Q is the charge on the surface, and r is the radius at which electric field is measured.

Calculate Charge on the Inner Surface

Given the magnitude of the electric field at the inner surface \(E_i = 80.0 \, N/C\), and the inner radius \(r_i = 8.00 \, cm\). Using the formula we derived in step 1, we can calculate the charge on the inner surface (\(Q_i\)): \(E_i = \dfrac{kQ_i}{r_i^2}\) \(q_i = \dfrac{E_ir_i^2}{k}\) We can convert the inner radius from cm to meters, \(r_i = 0.08\,m\), and plug in the values: \(Q_i = \dfrac{80 \, N/C × (0.08\,m)^2}{8.99 × 10^9 \, Nm^2/C^2} \) Calculating the value, we get \(Q_i \approx -2.26 × 10^{-11} C.\) The negative sign indicates that the charge on the inner surface is negative.

Calculate Charge on the Outer Surface

Given the magnitude of the electric field at the outer surface \(E_o = 80.0 \, N/C\), and the outer radius \(r_o = 10.0 \, cm\). Similarly, we can calculate the charge on the outer surface (\(Q_o\)): \(E_o = \dfrac{kQ_o}{r_o^2}\) \(Q_o = \dfrac{E_or_o^2}{k}\) We can convert the outer radius from cm to meters, \(r_o = 0.1\,m\), and plug in the values: \(Q_o = \dfrac{80 \, N/C × (0.1\,m)^2}{8.99 × 10^9 \, Nm^2/C^2} \) Calculating the value, we get \(Q_o \approx 8.90 × 10^{-11} C.\) Since there is no negative sign, the charge on the outer surface is positive.

Results

We have found the charges on the inner and outer surfaces of the spherical shell. The magnitude of the charge on the inner surface is approximately \(2.26 × 10^{-11} C\), and the magnitude of the charge on the outer surface is approximately \(8.90 × 10^{-11} C\). Notice that the charges on the inner and outer surfaces have opposite signs.

Key concepts

Electrostatics and the Behavior of Charge

Electrostatics is the branch of physics that deals with the study of stationary or slow-moving electric charges. When we explore electrostatics, we delve into the interactions between charged particles that are not moving or are moving at a very slow pace, relative to the speed of light. These interactions are governed by the fundamental principle known as Coulomb's law, which states that like charges repel each other, while opposite charges attract.

The electric charge is a fundamental property of matter carried by some subatomic particles. Charges come in two types, positive and negative, and the force between charges decreases with the square of the distance between them. In the context of the exercise, dealing with a conducting spherical shell, the electric charge distribution is uniform over the conductive material. A crucial point to understand is that while the entire shell may be charged, the electric field inside a hollow conductor in electrostatic equilibrium is zero. This phenomenon is explained by the fact that the charges on the conductor rearrange themselves to cancel any electric field within the hollow cavity of the sphere.

Gauss's Law and Its Applications

Gauss's law is a powerful tool in electrostatics, emerging from one of Maxwell's equations, which are the foundation of classical electromagnetism. It states that the electric flux through a closed surface is proportional to the enclosed electric charge. Mathematically, Gauss's law is represented as:
\[\Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0}\] where \(\Phi_E\) is the electric flux through a closed surface, \(Q_{\text{enc}}\) is the total charge enclosed by the surface, and \(\varepsilon_0\) is the vacuum permittivity.

When dealing with symmetrical charge distributions such as spherical shells, Gauss's law simplifies the process of calculating electric fields. In the context of the exercise, we can apply Gauss's law to a spherical surface concentric with the shell and use the symmetry of the problem to deduce the field only depends on the radial distance from the center, and not on the angle or direction. This allows us to infer that the charge inside a hollow conducting spherical shell does not affect the electric field outside the shell. Additionally, any excess charge on a conductor resides entirely on its outer surface, which is reflected in the solution's outcome, indicating a charge presence only on the sphere's exterior.

Understanding Electric Charge Quantification

The quantification of electric charge in the exercise uses a fundamental constant of electrostatics, the electrostatic constant \(k\), also known as Coulomb's constant. This constant, \(k = 8.99 \times 10^9 \, Nm^2/C^2\), forms part of Coulomb's law that calculates the electric force between two point charges:
\[F = k\frac{\lvert q_1q_2 \rvert}{r^2}\]

Applying this principle, we can determine the charge on the surfaces of a spherical shell by knowing the electric field and the radius. The mathematical relationship used in the solution is a rearrangement of Coulomb's law, which links the electric field (E), the charge (Q), and the radius (r) as in the formula \(E = \frac{kQ}{r^2}\). By rearranging it to solve for charge, \(Q = \frac{Er^2}{k}\), we determine the magnitude of charge by simply plugging in the values of the electric field and radius at a given point. Note the importance of converting units, such as radii from centimeters to meters before calculation to maintain consistency with SI units. From the exercise, the differing signs of the charges on the inner and outer surfaces reflect the directional nature of electric fields, pointing out that the inner surface charge is negative while the outer is positive.