### **Definition of Parameters** - **Surface charge density (\(\sigma\))**: \(-1.50 \times 10^{-6} \, \text{C/m}^2\) - **Initial velocity of proton (\(v_0\))**: \(1.80 \times 10^6 \, \text{m/s}\) - **Proton charge (\(q\))**: \(1.60 \times 10^{-19} \, \text{C}\) - **Proton mass (\(m\))**: \(1.67 \times 10^{-27} \, \text{kg}\) - **Permittivity of free space (\(\epsilon_0\))**: \(8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2\) --- ### **Step 1: Find the Electric Field due to the Plane** The electric field (\(E\)) near an infinite charged plane is given by: \[ E = \frac{\sigma}{2\epsilon_0} \] **Explanation:** The formula for the electric field due to an infinite plane is derived from Gauss's Law. The field points away from the plane if \(\sigma > 0\) and towards the plane if \(\sigma < 0\). Here, the field points **towards** the plane because \(\sigma\) is negative. \[ E = \frac{-1.50 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \] \[ E = \frac{-1.50 \times 10^{-6}}{1.77 \times 10^{-11}} \] \[ E = -8.47 \times 10^{4} \, \text{N/C} \] --- ### **Step 2: Use Energy Conservation to Find the Turning Point** The work done by the electric field stops the proton and brings it to rest at its maximum distance (\(\Delta x\)) from the plane. Use energy conservation: \[ \frac{1}{2}mv_0^2 = qEd \] Since the field is **towards** the plane and the proton is moving **away**, the work done by the field is **negative**: \[ \frac{1}{2}mv_0^2 = -q E \Delta x \] \[ \Delta x = \frac{\frac{1}{2}mv_0^2}{-qE} \] Plug in the values: \[ \Delta x = \frac{0.5 \times 1.67 \times 10^{-27} \times (1.80 \times 10^{6})^2}{-1.60 \times 10^{-19} \times (-8.47 \times 10^{4})} \] Calculate numerator: \[ 0.5 \times 1.67 \times 10^{-27} \times 3.24 \times 10^{12} = 2.704 \times 10^{-15} \] Calculate denominator: \[ 1.60 \times 10^{-19} \times 8.47 \times 10^4 = 1.3552 \times 10^{-14} \] \[ \Delta x = \frac{2.704 \times 10^{-15}}{1.3552 \times 10^{-14}} \] \[ \Delta x = 0.200 \, \text{m} \] --- ### **Final Answer Summary** The proton travels **0.200 m** before reaching its turning point. --- **Explanation Block for Each Step:** - **Step 1:** Found the electric field using the formula for an infinite plane and substituted the values, noting the direction and effect of negative charge density. - **Step 2:** Applied conservation of energy to equate the proton's initial kinetic energy to the work done by the electric field, then solved for the distance traveled before the proton stops. --- **Final boxed answer:** \[ \boxed{0.200 \ \text{m}} \]