Let's analyze the problem step by step. ## **Given Data** - Four parallel plates with surface charge densities: - Plate 1: \( \sigma_1 = 5\sigma \) - Plate 2: \( \sigma_2 = 5\sigma \) - Plate 3: \( \sigma_3 = -3\sigma \) - Plate 4: \( \sigma_4 = 4\sigma \) - Points A, B, C, D, E are located as shown (A left of plate 1, E right of plate 4). - The electric field is positive in the positive x-direction. ## **Electric Field Due to a Single Infinite Plane** The electric field due to an infinite plane with charge density \( \sigma \) is: \[ E = \frac{\sigma}{2\epsilon_0} \] - **To the right** of the plane: field points away if \( \sigma > 0 \), toward if \( \sigma < 0 \) - **To the left** of the plane: field points toward if \( \sigma > 0 \), away if \( \sigma < 0 \) ## **Net Electric Field Calculation** At each region, sum the contributions of all plates to the **right** of the observation point. ### **Regions:** - **A:** Left of all plates - **B:** Between plate 1 and 2 - **C:** Between plate 2 and 3 - **D:** Between plate 3 and 4 - **E:** Right of all plates Let's calculate the net field at each point: --- ### **A (Left of Plate 1)** All plates are to the right. - For each plate, field at A points to the **left** for \( \sigma > 0 \), to the **right** for \( \sigma < 0 \). - Field at A (due to all plates): \[ E_A = -\frac{1}{2\epsilon_0}(\sigma_1 + \sigma_2 + \sigma_3 + \sigma_4) \] \[ = -\frac{1}{2\epsilon_0} (5\sigma + 5\sigma - 3\sigma + 4\sigma) = -\frac{1}{2\epsilon_0}(11\sigma) \] --- ### **B (Between Plate 1 and 2)** Plates 2, 3, 4 are to the right; Plate 1 is to the left. - Field at B: - Plate 1: right of point, so field is \( +\frac{\sigma_1}{2\epsilon_0} \) - Plates 2, 3, 4: to the right of point, so field is \( -\frac{\sigma}{2\epsilon_0} \) \[ E_B = +\frac{1}{2\epsilon_0}\sigma_1 - \frac{1}{2\epsilon_0}(\sigma_2 + \sigma_3 + \sigma_4) \] \[ = \frac{1}{2\epsilon_0}\left[5\sigma - (5\sigma - 3\sigma + 4\sigma)\right] \] \[ = \frac{1}{2\epsilon_0}(5\sigma - 6\sigma) = -\frac{1}{2\epsilon_0}\sigma \] --- ### **C (Between Plate 2 and 3)** Plates 3, 4 are to the right; Plates 1, 2 to the left. - Field at C: \[ E_C = +\frac{1}{2\epsilon_0}(\sigma_1 + \sigma_2) - \frac{1}{2\epsilon_0}(\sigma_3 + \sigma_4) \] \[ = \frac{1}{2\epsilon_0}[(5\sigma + 5\sigma) - (-3\sigma + 4\sigma)] \] \[ = \frac{1}{2\epsilon_0}(10\sigma - 1\sigma) \] \[ = \frac{1}{2\epsilon_0}(9\sigma) \] --- ### **D (Between Plate 3 and 4)** Plate 4 is to the right; Plates 1, 2, 3 to the left. - Field at D: \[ E_D = +\frac{1}{2\epsilon_0}(\sigma_1 + \sigma_2 + \sigma_3) - \frac{1}{2\epsilon_0}\sigma_4 \] \[ = \frac{1}{2\epsilon_0}[(5\sigma + 5\sigma - 3\sigma) - 4\sigma] \] \[ = \frac{1}{2\epsilon_0}(7\sigma - 4\sigma) = \frac{1}{2\epsilon_0}(3\sigma) \] --- ### **E (Right of Plate 4)** All plates are to the left. - Field at E: \[ E_E = +\frac{1}{2\epsilon_0}(\sigma_1 + \sigma_2 + \sigma_3 + \sigma_4) \] \[ = \frac{1}{2\epsilon_0}(11\sigma) \] --- ## **Summary Table** | Point | Electric Field \( E \) | |-------|------------------------| | A | \( -\frac{11\sigma}{2\epsilon_0} \) | | B | \( -\frac{\sigma}{2\epsilon_0} \) | | C | \( \frac{9\sigma}{2\epsilon_0} \) | | D | \( \frac{3\sigma}{2\epsilon_0} \) | | E | \( \frac{11\sigma}{2\epsilon_0} \) | - Negative sign: field points in the negative x-direction. - Positive sign: field points in the positive x-direction. --- ## **Final Answers** - **A:** \( E = -\frac{11\sigma}{2\epsilon_0} \) - **B:** \( E = -\frac{\sigma}{2\epsilon_0} \) - **C:** \( E = \frac{9\sigma}{2\epsilon_0} \) - **D:** \( E = \frac{3\sigma}{2\epsilon_0} \) - **E:** \( E = \frac{11\sigma}{2\epsilon_0} \)