Question

The figure below shows an electron at the origin that is released with initial speed vo = 3.1 106 m/s at an angle θ。= 45° between the plates of a parallel plate capacitor of plate separation D = 2.0 mm. If the potential difference between the plates is Δν 160 V, calculate the closest proximity, d, of the electron to the bottom plate (in mm) D 0 0o 0

Answer 1

V_0 = 3.1 times 10^6 m/s theta_0 = 45 degree D = 2.0 mm = 2.0 times 10^-3 m delta V = 160V The electric force F = qE = q delta V/D acceleration of electron a = F/m = q delta V/mD a = 1.6 times 10^-19 c times 160V/9.1 times 10^-31 kg times 2.0 times 10^-3 m a = 1.406 times 10^16 m/s^2 the vertical velocity of electron v_0y = V_0 sin theta_0 v_oy = 3.1 times 10^6 m/s times sin 45 degree v_0y = 2.192 times 10^6 m/s the minimum position y = v^2_0y y = (2.192 times 10^6 m/s)^2/ 2 times 1.406 times 10^16 m/s^2 y = 1.708 times 10^-4 m y = 1.708 times 10^-4 m y = 0.1708 33 d = D/2 - y d = 2.0 mm/2 - 0.1708 mm d = 0.8292 mm