A potassium ion with an initial velocity of 10^6 m/s in the x direction is shot

Question

A potassium ion with an initial velocity of 10^6 m/s in the x direction is shot through the center of two capacitor plates located at y = 8 cm and y = -8cm. The plate at y = 8 cm has a charge of -10 nC and the plate at y = -8 cm has a charge of +10 nC. The plates are circular with a radius of 3 cm. Neglecting gravity, what is the y position of the ion as it exits the capacitor? For this problem remember that: y_t = y_1 | v_y delta t | 1/2 B_y (delta t)^2 and x_t = x_1 | v_x delta t | 1/2 B_x (delta t)^2 See the diagram below.

Explanation / Answer

sigma = q / A = (10 x 10^-9) / (pi x 0.03^2)

sigma = 3.537 x 10^-6 C/m^2

E = sigma / e0 = 4 x 10^5 N/C

a = q E / m = (1.6 x 10^-19) (4 x 10^5) / (39 x 1.67 x 10^-27)

a = 9.813 x 10^11 m/s^2

t = (0.06) / (10^6) = 6 x 10^-8 sec

y = v0 t + a t^2 / 2

y = 0 + (9.813 x 10^11)(6 x 10^-8)^2 / 2

y = 1.77 x 10^-3 m Or 1.77 mm ........Ans

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