Here’s a **step-by-step outline** to solve the problem: --- ## **Problem Breakdown** - **Given:** - A narrow charged solid cylinder (inner, conducting) inside a larger charged cylindrical shell (outer, conducting). - Both have uniform surface charge **densities**. - The graph shows *E vs. r* (radial distance). - The vertical axis: \( E \) in \( 4.2 \times 10^2 \) N/C units. - We need to **find the linear charge density (\(\lambda\)) of the shell**. --- ## **Step 1: Understanding the Graph** - The graph shows \( E \) (electric field) as a function of \( r \) (distance from the axis). - At the location outside the shell (largest \( r \)), \( E \) drops to zero (as expected for a conducting shell). - The value of \( E \) just outside the inner cylinder (before the shell starts) is **2.8** units on the y-axis. --- ## **Step 2: Find the Actual Value of \( E \)** - The vertical axis scale is set so that 1 unit = \( 4.2 \times 10^2 \) N/C. - The value just outside the inner cylinder is 2.8 units, so: \[ E = 2.8 \times 4.2 \times 10^2 = 2.8 \times 420 = 1176~\text{N/C} \] --- ## **Step 3: Gauss's Law for Cylindrical Shell** For a long line of charge (or cylinder), the electric field at radius \( r \) outside the cylinder is: \[ E = \frac{\lambda}{2\pi \epsilon_ r} \] Where: - \( \lambda \) = linear charge density - \( r \) = radial distance from the axis - \( \epsilon_ \) = permittivity of free space, \( 8.85 \times 10^{-12}~\text{C}^2/\text{N}\cdot\text{m}^2 \) --- ## **Step 4: Find \( r \) at the Location of \( E \)** - On the graph, at \( r = 4 \) mm, \( E = 2.8 \) units. - So, \( r = 4 \) mm \( = 4 \times 10^{-3} \) m. --- ## **Step 5: Solve for \( \lambda \)** Plug values into the formula: \[ E = \frac{\lambda}{2\pi \epsilon_ r} \] \[ \lambda = E \cdot 2\pi \epsilon_ r \] Insert the numbers: - \( E = 1176~\text{N/C} \) - \( r = 4 \times 10^{-3}~\text{m} \) - \( \epsilon_ = 8.85 \times 10^{-12}~\text{C}^2/\text{N}\cdot\text{m}^2 \) \[ \lambda = 1176 \times 2\pi \times 8.85 \times 10^{-12} \times 4 \times 10^{-3} \] Calculate step by step: - \( 2\pi \approx 6.283 \) - \( 8.85 \times 4 = 35.4 \) - \( 35.4 \times 10^{-15} = 3.54 \times 10^{-14} \) - \( 6.283 \times 3.54 \approx 22.25 \) - \( 1176 \times 22.25 \approx 26160 \) - \( 26160 \times 10^{-14} = 2.616 \times 10^{-10} \) So, \[ \boxed{\lambda = 2.6 \times 10^{-10}~\text{C/m}} \] --- ### **Final Answer** **Linear charge density of the shell:** \[ \boxed{2.6 \times 10^{-10}~\text{C/m}} \] --- Let me know if you need a step explained in more detail!