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 3.9 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) At x = 3.5, the kinetic energy = 3.9 eV and the potential energy is - 4eV. So the mechanical energy at this point will be, ME = 3.9-4 = -0.1 eV.

At turning point the kinetic energy will be zero so the mechnical energy = potential energy at turning point. But since the net mechanical energy is conserved so the potential energy at turning point will be equal to -0.1 eV.

14-6x = -0.1 => x = 2.35 cm(turning point)

b) Here too, at tuning point the kinetic energy will be zero. So the potential energy will be -0.1 eV.

4x-24 = -0.1 => x = 5.975 cm.

c) and d)

At x = 3- or just left to x=3 , the force will be e times minus of slope at left to x = 3. The slope is -6 V/cm, so the force will be 6 eV/cm. The magnitude is 6eV/cm and the direction is +x direction.

e) and f)

At just right to x = 5 cm, the slope is 4 v/cm so the magnitude of force will be 4eV/cm and the direction is - x direction.

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