Proton in a well. Figure 24-54 shows electric potential V along an x axis. A pro

Question

Proton in a well. Figure 24-54 shows electric potential V along an x axis. A proton is to be released at x = 3.5 cm with initial kinetic energy 3.6 eV. The scale of the vertical axis is set by Vs = 10.0 V. (a) If it is initially moving in the negative direction, it either reaches a turning point (if so, what is the x coordinate of that point) or it escapes from the plotted region (if so, what is its speed at x = 0)? (b) If it is initially moving in the positive direction, it either reaches a turning point (if so, what is the x coordinate of that point) or it escapes from the plotted region (if so, what is its speed at x - 6.0 cm)? (c) What is the electric force (including sign) on the proton if the proton moves just to the left of x = 3.0 cm? (d) What is the electric force (including sign) on the proton if the proton moves just to the right of x = 5.0 cm?

Explanation / Answer

a)

Voltage between 1cm and 3cm:

V = 9 - (6/2) (x-1) = 9 + 3 - 3 x = 12 - 3 x

K = 3.6 eV

Einitial = 3.6 + 3 = 6.6 eV

12 - 3 x = 6.6

x = 1.8 cm

b)

It will escape from the plotted region

Einitial = 3.6 + 3 = 6.6 eV

Vfinal = 5

Kfinal = 6.6 - 5 = 1.6 eV

0.5 m v^2 = K

v = (2K/m) = (2*1.6*1.6e-19/1.673e-27)

v = 1.75e4 m/s

c)

V = 9 - (6/2) (x-1) = 9 + 3 - 3 x = 12 - 300 x

U = q V = 1.6e-19 * (12 - 300 x)

F = - dU/dx = 1.6e-19*300 = 4.8e-17 N

d)

V = 6 + (4/1) (x-5) = -7 + 200 x

U = q V = 1.6e-19 * (-7 + 200 x)

F = - dU/dx = 1.6e-19*200 = -3.2e-17 N

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