Proton in a well. The figure shows electric potential V along an x axis. A proto

Question

Proton in a well. The figure shows electric potential V along an x axis. A proton is to be released at x = 3.5 cm with initial kinetic energy 3.8 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)? What are the (c) magnitude F and (d) direction (positive or negative direction of the x axis) of the electric force on the proton if the proton moves just to the left of x=3.0 cm? What are (e) F and (f) the direction if the proton moves just to the right of x=5.0 cm?

Explanation / Answer

(a) When the proton is released, its energy is K + U = 3.8 eV + 3.0 eV (the latter value is inferred from the graph). This implies that if we draw a horizontal line at the 6.8 Volt “height” in the graph and find where it intersects the voltage plot, then we can determine the turning point. Interpolating in the region between 1.0 cm and 3.0 cm, we find the turning point is at roughly x = 1.8 cm.

(b) There is no turning point towards the right, so the speed there is nonzero, and is given by energy conservation:

v=sqrt[(2 x 6.8eV)/m] = sqrt [(2 x 6.8 x 1.6 x 10-19/1.67 x 10-27)]

v = = = 36.1 km/s.

(c) The electric field at any point P is the (negative of the) slope of the voltage graph evaluated at P. Once we know the electric field, the force on the proton follows immediately from F= qE, where q = +e for the proton. In the region just to the left of x = 3.0 cm, the field is = (+300 V/m) (i cap unit vector) and the force is F = +4.8 x10-17 N.

(d) The force F points in the +x direction, as the electric field E.

(e) In the region just to the right of x = 5.0 cm, the field is =(–200 V/m) (i cap unit vector) and the magnitude of the force is F = 3.2 x 10-17 N.

(f) The force F points in the -x direction, as the electric field E.

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