Let's break down the problem **step by step**: ### Given: - Three concentric metallic shells. - Inner radii: \( R, 2R, 3R \) - Thickness of each shell: \( d = 0.1R \) - Charges: First shell = \( Q \), Second shell = \( -Q \), Third shell = \( 3Q \) - Find the magnitude of the electric field at \( r = 2.05R \). - Use \( k = \frac{1}{4\pi\epsilon_0} \). --- ## **Step 1: Identify the Location** At \( r = 2.05R \): - \( R < 2.05R < 2R+0.1R = 2.1R \) - So, **inside the second shell** (after passing the first shell, inside the body of the second shell). --- ## **Step 2: Gauss's Law** The **electric field** at a distance \( r \) from the center (inside a shell or between shells) is determined **only by the total charge enclosed within \( r \)**. \[ E \cdot 4 \pi r^2 = \frac{Q_{\text{enc}}}{\epsilon_0} \] \[ E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q_{\text{enc}}}{r^2} = k \frac{Q_{\text{enc}}}{r^2} \] --- ## **Step 3: Calculate \( Q_{\text{enc}} \) at \( r = 2.05R \)** - **First shell (radius R; thickness 0.1R):** Inner radius = \(R\), outer radius = \(1.1R\). Charge = \(Q\). - **Second shell (radius 2R; thickness 0.1R):** Inner radius = \(2R\), outer radius = \(2.1R\). Charge = \(-Q\). - \( 2.05R \) is between \(2R\) and \(2.1R\), **inside the conducting material** of the second shell. #### **What is the enclosed charge?** - Any Gaussian surface at \( r = 2.05R \) will enclose: - Charge on the first shell: \( Q \) - **None of the charge on the second and third shells, since their charge is on their surfaces** (by electrostatics, charge resides on the surface of conductors). Thus, **enclosed charge \( Q_{\text{enc}} = Q \)** --- ## **Step 4: Electric Field at \( r = 2.05R \) (inside a conductor)** BUT: Inside the conducting material of a shell, **the electric field is zero** (because the charges rearrange themselves so that the field inside the conductor cancels out). So, **at \( r = 2.05R \), which is inside the conducting material of the second shell, the electric field is zero**. --- ## **Step 5: Final Answer** \[ \boxed{E(r = 2.05R) = 0} \] **Explanation**: - The point \( r = 2.05R \) lies within the conducting material of the second shell. - Inside a conductor in electrostatic equilibrium, the electric field is always zero. --- ### **Summary Table** | Region | Range | Enclosed Charge | Electric Field | |---------------|----------------------|---------------------|-------------------------------| | Inside 1st shell | \( r < R \) | 0 | 0 | | Between 1st and 2nd shell | \( R < r < 1.1R \) | \( Q \) | \( k \frac{Q}{r^2} \) | | Between shells | \( 1.1R < r < 2R \) | \( Q \) | \( k \frac{Q}{r^2} \) | | **Inside 2nd shell (conductor)** | \( 2R < r < 2.1R \) | \( Q \) | **0** | | Between 2nd and 3rd shell | \( 2.1R < r < 3R \) | \( Q - Q = 0 \) | 0 | | Inside 3rd shell (conductor) | \( 3R < r < 3.1R \) | \( Q - Q = 0 \) | 0 | --- ## **Boxed Final Answer** \[ \boxed{ E(r = 2.05R) = 0 } \] If you need the electric field at a different region, let me know!