Let's solve each part step by step. Given: - Metal sphere of radius \(1\, \text{m}\), charge \(+5\, \text{C}\) - Concentric metal shell: inner radius \(3\, \text{m}\), outer radius \(4\, \text{m}\) - Shell carries **no net charge** ### (a) Find the electric field **E** at regions **A, B, and D**. Let: - **Region A:** \(r < 1\, \text{m}\) (inside the metal sphere) - **Region B:** \(1 < r < 3\, \text{m}\) (between sphere and shell) - **Region D:** \(r > 4\, \text{m}\) (outside the shell) #### **Step 1: Use Gauss’s Law for each region** - **Region A** (\(r < 1\, \text{m}\), inside the conductor): **Explanation:** In electrostatic equilibrium, the electric field inside a conductor is **zero**. \[ E_{\text{A}} = \] - **Region B** (\(1 < r < 3\, \text{m}\), between sphere and shell): **Explanation:** The total charge enclosed by a Gaussian surface of radius \(r\) is \(5\, \text{C}\). By Gauss’s Law: \[ \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\epsilon_} \implies E \cdot 4\pi r^2 = \frac{5}{\epsilon_} \] \[ E_{\text{B}} = \frac{1}{4\pi \epsilon_} \frac{5}{r^2} \] #### **Step 2: Continue for Region D** - **Region D** (\(r > 4\, \text{m}\), outside the shell): **Explanation:** The shell carries no net charge, so the total enclosed charge is still \(5\, \text{C}\). By Gauss’s Law (same as above): \[ E_{\text{D}} = \frac{1}{4\pi \epsilon_} \frac{5}{r^2} \] --- ### **Final Answers (a):** - **Region A:** \(E = \) - **Region B:** \(E = \frac{1}{4\pi \epsilon_} \frac{5}{r^2}\) - **Region D:** \(E = \frac{1}{4\pi \epsilon_} \frac{5}{r^2}\) --- Would you like to proceed to part (b)?

## Electric Field Analysis of a Charged Sphere Given: - Metal sphere of radius \(1\, \text{m}\), charge \(+5\, \text{C}\) - Concentric metal shell with inner radius \(3\, \text{m}\) and outer radius \(4\, \text{m}\) - Shell carries no net charge ### (a) Find the Electric Field **E** at Regions **A**, **B**, and **D**. #### **Step 1: Calculate Electric Field in Regions A and B** - **Region A** (\(r < 1\, \text{m}\)): **Explanation:** Inside a conductor, the electric field is zero due to the redistribution of charges. \[ E_{\text{A}} = 0 \] - **Region B** (\(1 < r < 3\, \text{m}\)): **Explanation:** The electric field can be determined using Gauss's Law, where the charge enclosed is \(+5\, \text{C}\). The integral form of Gauss's Law is: \[ \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\epsilon_0} \] For a spherical surface of radius \(r\): \[ E \cdot 4\pi r^2 = \frac{5}{\epsilon_0} \implies E_{\text{B}} = \frac{5}{4\pi \epsilon_0 r^2} \] #### **Step 2: Calculate Electric Field in Region D** - **Region D** (\(r > 4\, \text{m}\)): **Explanation:** Outside the shell, the charge enclosed remains \(+5\, \text{C}\), thus the electric field behaves similarly to that in Region B. Using Gauss's Law: \[ E_{\text{D}} = \frac{5}{4\pi \epsilon_0 r^2} \] ### **Final Answers (a):** - **Region A:** \(E_{\text{A}} = 0\) - **Region B:** \(E_{\text{B}} = \frac{5}{4\pi \epsilon_0 r^2}\) - **Region D:** \(E_{\text{D}} = \frac{5}{4\pi \epsilon_0 r^2}\) --- ### (b) Find the Surface Charge Density on the Inner and Outer Surfaces #### **Step 1: Calculate Surface Charge Density on Inner Surface** - **Inner surface (at \(r = 3\, \text{m}\))**: **Explanation:** The inner surface of the shell will have an induced charge of \(-5\, \text{C}\) to maintain electrostatic equilibrium (screening the inner charge). The area of the inner surface: \[ A_{\text{inner}} = 4\pi (3\, \text{m})^2 = 36\pi\, \text{m}^2 \] The surface charge density (\(\sigma_{\text{inner}}\)) is: \[ \sigma_{\text{inner}} = \frac{-5\, \text{C}}{36\pi\, \text{m}^2} \] #### **Step 2: Calculate Surface Charge Density on Outer Surface** - **Outer surface (at \(r = 4\, \text{m}\))**: **Explanation:** The outer surface will have a charge of \(0\, \text{C}\) since the shell has no net charge. Thus: \[ \sigma_{\text{outer}} = 0 \] ### **Final Answers (b):** - **Inner Surface Charge Density:** \(\sigma_{\text{inner}} = \frac{-5}{36\pi}\, \text{C/m}^2\) - **Outer Surface Charge Density:** \(\sigma_{\text{outer}} = 0\, \text{C/m}^2\) --- ### (c) Find the Potential at the Sphere's Center #### **Step 1: Determine the Electric Potential** - **At the center (Region A)**: **Explanation:** The potential inside a conductor is constant and equals the potential at the surface of the conductor. The potential \(V\) at the surface (\(r = 1\, \text{m}\)) can be calculated using: \[ V = k \cdot \frac{q}{r} = \frac{1}{4\pi \epsilon_0} \cdot \frac{5}{1} = \frac{5}{4\pi \epsilon_0} \] ### **Final Answer (c):** - **Potential at the Sphere's Center:** \(V = \frac{5}{4\pi \epsilon_0}\) --- Would you like further assistance with any other part of this problem?