Let's solve the problem **step-by-step**: ### **Problem Restatement** - Two concentric spherical shells. - Surface charge densities: \( \sigma_1 \) (on the inner shell), \( \sigma_2 \) (on the outer shell). - Find the electric field **between the shells** using Gauss's Law. --- ### **Step 1: Assign Radii** - Let the radius of the inner shell be \( R_1 \). - Let the radius of the outer shell be \( R_2 \). - So, the region **between the shells** is: \( R_1 < r < R_2 \). --- ### **Step 2: Total Charge on Inner Shell** - Surface charge density = charge per unit area. - Area of inner shell = \( 4\pi R_1^2 \). - Total charge on inner shell: \( Q_1 = \sigma_1 \cdot 4\pi R_1^2 \). --- ### **Step 3: Apply Gauss's Law** Gauss's Law: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0} \] - Choose a Gaussian surface: a sphere of radius \( r \) such that \( R_1 < r < R_2 \). - The electric field is **radial** and **constant** over the surface. So, \[ E \cdot 4\pi r^2 = \frac{Q_{\text{enclosed}}}{\epsilon_0} \] --- ### **Step 4: Enclosed Charge** - Only the **inner shell's charge** is enclosed (since we are between the shells). - Enclosed charge: \( Q_{\text{enclosed}} = Q_1 = \sigma_1 \cdot 4\pi R_1^2 \) --- ### **Step 5: Solve for \( E \)** Plug in values: \[ E \cdot 4\pi r^2 = \frac{\sigma_1 \cdot 4\pi R_1^2}{\epsilon_0} \] \[ E = \frac{\sigma_1 R_1^2}{\epsilon_0 r^2} \] --- ### **Step 6: Final Answer** \[ \boxed{ E(r) = \frac{\sigma_1 R_1^2}{\epsilon_0 r^2} \qquad (R_1 < r < R_2) } \] **Note:** - The charge on the outer shell (\( \sigma_2 \)) does **not** contribute to the field between the shells because it is outside the Gaussian surface. - If you need the field **outside both shells** (\( r > R_2 \)), then you add both charges. --- Let me know if you need the field **inside** the inner shell or **outside** the outer shell!