Let's solve the problem step by step: --- ## **Given Data** - \( Q_1 = +12\,\mu\text{C} = 12 \times 10^{-6}\,\text{C} \) at \( x_1 = -.2\,\text{m} \) - \( Q_2 = -8\,\mu\text{C} = -8 \times 10^{-6}\,\text{C} \) at \( x_2 = +.3\,\text{m} \) - \( m_p = 1.67 \times 10^{-27}\,\text{kg} \) (mass of proton) - \( e = 1.6 \times 10^{-19}\,\text{C} \) (charge of proton) - \( k = 8.99 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2 \) (Coulomb's constant) --- ## **(a) Electric Field Zero Point(s)** ### **Electric Field Due to Point Charges** The net electric field at point \( x \) on the x-axis: \[ E = E_1 + E_2 \] Where: - \( E_1 = \frac{kQ_1}{(x-x_1)^2} \) (direction: away from \( Q_1 \)) - \( E_2 = \frac{kQ_2}{(x-x_2)^2} \) (direction: toward \( Q_2 \), since it's negative) Set \( E = \): \[ \frac{kQ_1}{(x+.2)^2} + \frac{kQ_2}{(x-.3)^2} = \] \[ \frac{Q_1}{(x+.2)^2} + \frac{Q_2}{(x-.3)^2} = \] \[ \frac{Q_1}{(x+.2)^2} = -\frac{Q_2}{(x-.3)^2} \] Plug in values (\( Q_1 > , Q_2 < \)): \[ \frac{12}{(x+.2)^2} = \frac{8}{(x-.3)^2} \] \[ \frac{12}{(x+.2)^2} - \frac{8}{(x-.3)^2} = \] \[ \frac{12}{(x+.2)^2} = \frac{8}{(x-.3)^2} \] Cross-multiplied: \[ 12(x-.3)^2 = 8(x+.2)^2 \] \[ 12(x^2 - .6x + .09) = 8(x^2 + .4x + .04) \] \[ 12x^2 - 7.2x + 1.08 = 8x^2 + 3.2x + .32 \] \[ (12x^2 - 8x^2) + (-7.2x - 3.2x) + (1.08 - .32) = \] \[ 4x^2 - 10.4x + .76 = \] ### **Quadratic Equation** \[ 4x^2 - 10.4x + .76 = \] \[ x = \frac{10.4 \pm \sqrt{(-10.4)^2 - 4 \times 4 \times .76}}{2 \times 4} \] \[ x = \frac{10.4 \pm \sqrt{108.16 - 12.16}}{8} \] \[ x = \frac{10.4 \pm \sqrt{96}}{8} \] \[ \sqrt{96} \approx 9.798 \] \[ x_1 = \frac{10.4 + 9.798}{8} = \frac{20.198}{8} \approx 2.525 \] \[ x_2 = \frac{10.4 - 9.798}{8} = \frac{.602}{8} \approx .075 \] ### **Check Validity** - \( x_1 = 2.525\,\text{m} \) (far to the right, possible) - \( x_2 = .075\,\text{m} \) (between the charges, but both fields point in the same direction, so not valid **between** them). #### **Final Answer (a):** **The electric field is zero at \( x \approx 2.53\,\text{m} \) on the x-axis.** --- ## **(b) Electric Potential Zero Point(s)** Electric potential at \( x \): \[ V = V_1 + V_2 = \frac{kQ_1}{x+.2} + \frac{kQ_2}{x-.3} \] Set \( V = \): \[ \frac{Q_1}{x+.2} + \frac{Q_2}{x-.3} = \] \[ \frac{12}{x+.2} + \frac{-8}{x-.3} = \] \[ \frac{12}{x+.2} = \frac{8}{x-.3} \] \[ 12(x-.3) = 8(x+.2) \] \[ 12x - 3.6 = 8x + 1.6 \] \[ 12x - 8x = 1.6 + 3.6 \] \[ 4x = 5.2 \] \[ x = 1.3 \] #### **Final Answer (b):** **The electric potential is zero at \( x = 1.3\,\text{m} \) on the x-axis.** --- ## **(c) Speed of Proton Released from Rest at \( x = .1\,\text{m} \), Reaches \( x = .5\,\text{m} \)** ### **Conservation of Energy** \[ K_i + U_i = K_f + U_f \] Initial: \( K_i = \) (released from rest) \[ + U_i = \frac{1}{2}mv^2 + U_f \] \[ U_i - U_f = \frac{1}{2}mv^2 \] \[ v = \sqrt{\frac{2 (U_i - U_f)}{m}} \] Potential energy for charge \( q \) at position \( x \): \[ U = qV(x) \] \[ U_i = e \left[ \frac{kQ_1}{x_i + .2} + \frac{kQ_2}{x_i - .3} \right] \] \[ U_f = e \left[ \frac{kQ_1}{x_f + .2} + \frac{kQ_2}{x_f - .3} \right] \] #### **Calculate \( V(x_i) \) and \( V(x_f) \):** - \( x_i = .1 \) - \( x_f = .5 \) \[ V(x_i) = \frac{kQ_1}{.1 + .2} + \frac{kQ_2}{.1 - .3} \] \[ = \frac{8.99 \times 10^9 \times 12 \times 10^{-6}}{.3} + \frac{8.99 \times 10^9 \times (-8) \times 10^{-6}}{-.2} \] \[ = \frac{8.99 \times 10^9 \times 12 \times 10^{-6}}{.3} + \frac{8.99 \times 10^9 \times 8 \times 10^{-6}}{.2} \] Calculate each: - \( \frac{8.99 \times 10^9 \times 12 \times 10^{-6}}{.3} = \frac{8.99 \times 12 \times 10^3}{.3} \) - \( 8.99 \times 12 = 107.88 \) - \( \frac{107.88 \times 10^3}{.3} = 359,600 \) - \( \frac{8.99 \times 8 \times 10^3}{.2} = \frac{71.92 \times 10^3}{.2} = 359,600 \) \[ V(x_i) = 359,600 + 359,600 = 719,200\,\text{V} \] #### **At \( x_f = .5 \):** - \( x_f + .2 = .7 \) - \( x_f - .3 = .2 \) \[ V(x_f) = \frac{8.99 \times 10^9 \times 12 \times 10^{-6}}{.7} + \frac{8.99 \times 10^9 \times (-8) \times 10^{-6}}{.2} \] \[ = \frac{107.88 \times 10^3}{.7} - \frac{71.92 \times 10^3}{.2} \] \[ = 154,114 - 359,600 = -205,486\,\text{V} \] #### **Potential Difference:** \[ \Delta V = V(x_i) - V(x_f) = 719,200 - (-205,486) = 924,686\,\text{V} \] #### **Kinetic Energy:** \[ \frac{1}{2}mv^2 = e\Delta V \] \[ v = \sqrt{\frac{2e\Delta V}{m}} \] \[ = \sqrt{ \frac{2 \times 1.6 \times 10^{-19} \times 924,686}{1.67 \times 10^{-27}} } \] \[ = \sqrt{ \frac{2 \times 1.6 \times 10^{-19} \times 924,686}{1.67 \times 10^{-27}} } \] \[ 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \] \[ 3.2 \times 10^{-19} \times 924,686 = 2,959,000 \times 10^{-19} \] \[ = 2.959 \times 10^{-13} \] \[ v = \sqrt{ \frac{2.959 \times 10^{-13}}{1.67 \times 10^{-27}} } \] \[ = \sqrt{1.773 \times 10^{14}} \] \[ = 1.33 \times 10^7\,\text{m/s} \] --- ## **Final Answers** ### **(a) Electric field is zero at:** \[ \boxed{x \approx 2.53\,\text{m}} \] ### **(b) Electric potential is zero at:** \[ \boxed{x = 1.3\,\text{m}} \] ### **(c) Speed of the proton at \( x = .5\,\text{m} \):** \[ \boxed{v \approx 1.33 \times 10^7\,\text{m/s}} \]