Let's break down and solve the problem step by step: ### **Problem Recap** - Two point charges: \( -2q \) at distance \( d \) and \( q \) at distance \( 3d \) from the center of a hollow, grounded, conducting sphere of inner radius \( a \). - **Goal:** Using the **method of images**, find the **electric force on charge \( q \)**. --- ## **Step 1: Setup and Image Charges** When a point charge is placed outside a grounded conducting sphere, it induces an image charge inside the sphere. For a charge \( Q \) at distance \( r \) from the center of a grounded sphere of radius \( a \): - **Image charge \( Q' \):** \[ Q' = -Q \frac{a}{r} \] - **Image charge position:** At distance \( r' = \frac{a^2}{r} \) from the center, along the line joining the center and the real charge. ### **For This Problem** We have **two real charges**: 1. \( -2q \) at \( r_1 = d \) 2. \( q \) at \( r_2 = 3d \) #### **Image Charges:** - For \( -2q \) at \( d \): - Image charge: \( Q_1' = 2q \frac{a}{d} \) - Image position: \( r_1' = \frac{a^2}{d} \) - For \( q \) at \( 3d \): - Image charge: \( Q_2' = -q \frac{a}{3d} \) - Image position: \( r_2' = \frac{a^2}{3d} \) --- ## **Step 2: Forces Acting on Charge \( q \) at \( 3d \)** The **total force on \( q \) at \( 3d \)** is the sum of the forces from: 1. The real charge \( -2q \) at \( d \) 2. The image charge \( Q_1' \) at \( r_1' \) 3. The image charge \( Q_2' \) at \( r_2' \) (Remember, the image charge of \( q \) is only a mathematical device for boundary conditions and does **not** exert a force on its own real charge.) --- ### 1. **Force due to real charge \( -2q \) at \( d \):** \[ F_1 = \frac{1}{4\pi\epsilon_0} \frac{q \cdot (-2q)}{(3d - d)^2} = -\frac{2q^2}{4\pi\epsilon_0 (2d)^2} = -\frac{q^2}{8\pi\epsilon_0 d^2} \] --- ### 2. **Force due to image of \( -2q \): \( Q_1' = 2q \frac{a}{d} \) at \( r_1' = \frac{a^2}{d} \):** Separation between \( q \) at \( 3d \) and image at \( a^2/d \): \[ R_2 = 3d - \frac{a^2}{d} \] \[ F_2 = \frac{1}{4\pi\epsilon_0} \frac{q \cdot 2q \frac{a}{d}}{\left(3d - \frac{a^2}{d}\right)^2} = \frac{2q^2 a/d}{4\pi\epsilon_0 \left(3d - \frac{a^2}{d}\right)^2} = \frac{2q^2 a}{4\pi\epsilon_0 d \left(3d - \frac{a^2}{d}\right)^2} \] --- ### 3. **Force due to image of \( q \): \( Q_2' = -q \frac{a}{3d} \) at \( r_2' = \frac{a^2}{3d} \):** Separation between \( q \) at \( 3d \) and its image at \( a^2/(3d) \): \[ R_3 = 3d - \frac{a^2}{3d} \] \[ F_3 = \frac{1}{4\pi\epsilon_0} \frac{q \cdot \left(-q \frac{a}{3d}\right)}{\left(3d - \frac{a^2}{3d}\right)^2} = -\frac{q^2 a}{12\pi\epsilon_0 d \left(3d - \frac{a^2}{3d}\right)^2} \] --- ## **Step 3: Add Up the Forces** Total force on \( q \): \[ \boxed{ F_{\text{total}} = F_1 + F_2 + F_3 } \] Substitute the values: \[ F_{\text{total}} = -\frac{q^2}{8\pi\epsilon_0 d^2} + \frac{2q^2 a}{4\pi\epsilon_0 d \left(3d - \frac{a^2}{d}\right)^2} - \frac{q^2 a}{12\pi\epsilon_0 d \left(3d - \frac{a^2}{3d}\right)^2} \] Or, simplifying the coefficients: \[ \boxed{ F_{\text{total}} = -\frac{q^2}{8\pi\epsilon_0 d^2} + \frac{q^2 a}{2\pi\epsilon_0 d \left(3d - \frac{a^2}{d}\right)^2} - \frac{q^2 a}{12\pi\epsilon_0 d \left(3d - \frac{a^2}{3d}\right)^2} } \] --- ### **Final Answer** \[ \boxed{ \text{The electric force acting on charge } q \text{ is:} } \] \[ \boxed{ F = -\frac{q^2}{8\pi\epsilon_0 d^2} + \frac{q^2 a}{2\pi\epsilon_0 d \left(3d - \frac{a^2}{d}\right)^2} - \frac{q^2 a}{12\pi\epsilon_0 d \left(3d - \frac{a^2}{3d}\right)^2} } \] **Direction:** Along the line joining the charges and the center (radially outward/inward depending on sign).