Charged Particle Deflection E60361

1. **State the problem:** Calculate the downward deflection of a negatively charged particle moving horizontally between parallel plates with a given potential difference, plate separation, and plate length, considering gravity. 2. **Given data:** - Charge, $q = 6.40 \times 10^{-19}$ C - Mass, $m = 1.50 \times 10^{-12}$ kg - Initial horizontal velocity, $v_x = 150$ m/s - Potential difference, $V = 25000$ V - Plate separation, $d = 5.00$ cm = $0.0500$ m - Plate length, $L = 10.0$ cm = $0.100$ m 3. **Find:** Downward deflection $y$ when the particle leaves the plates. 4. **Formulas and concepts:** - Electric field between plates: $$E = \frac{V}{d}$$ - Force on particle due to electric field: $$F_e = qE$$ - Acceleration due to electric force: $$a_e = \frac{F_e}{m} = \frac{qE}{m}$$ - Acceleration due to gravity: $$a_g = 9.8 \text{ m/s}^2$$ downward - Total downward acceleration: $$a_y = a_e + a_g$$ (both downward) - Time to cross plates horizontally: $$t = \frac{L}{v_x}$$ - Downward deflection: $$y = \frac{1}{2} a_y t^2$$ 5. **Calculate electric field:** $$E = \frac{25000}{0.0500} = 500000 \text{ V/m}$$ 6. **Calculate electric force:** $$F_e = (6.40 \times 10^{-19})(500000) = 3.20 \times 10^{-13} \text{ N}$$ 7. **Calculate electric acceleration:** $$a_e = \frac{3.20 \times 10^{-13}}{1.50 \times 10^{-12}} = 0.2133 \text{ m/s}^2$$ 8. **Calculate total downward acceleration:** $$a_y = 0.2133 + 9.8 = 10.0133 \text{ m/s}^2$$ 9. **Calculate time to cross plates:** $$t = \frac{0.100}{150} = 6.67 \times 10^{-4} \text{ s}$$ 10. **Calculate downward deflection:** $$y = \frac{1}{2} (10.0133) (6.67 \times 10^{-4})^2$$ $$y = 0.5 \times 10.0133 \times 4.44 \times 10^{-7}$$ $$y = 2.22 \times 10^{-6} \text{ m}$$ **Final answer:** The downward deflection of the particle when it leaves the plates is approximately $$2.22 \times 10^{-6} \text{ meters}$$ or $$2.22 \text{ micrometers}$$.