The figure below shows electric potential V along an x axis. The scale of the ve
ID: 1530320 • Letter: T
Question
The figure below shows electric potential V along an x axis. The scale of the vertical axis is set by Vs = 16.0 V. A proton is to be released at x = 3.5 cm with initial kinetic energy 4.68 eV.
(a) If it is initially moving in the negative direction of the axis, does it reach a turning point (if so, what is the xcoordinate of that point) or does it escape from the plotted region (if so, what is its speed at x = 0)?
____________ select cm or m/s
(b) If it is initially moving in the positive direction of the axis, does it reach a turning point (if so, what is the xcoordinate of that point) or does it escape from the plotted region (if so, what is its speed at x = 6.0 cm)?
____________ select cm or m/s
(c) What is the magnitude F of the electric force on the proton if the proton moves just to the left of x = 3.0 cm?
____________N
(d) What is the 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?
positive x direction
(e) What is F if the proton moves just to the right of x = 5.0 cm? (Enter the magnitude.)
____________N
(f) What is the direction if the proton moves just to the right of x = 5.0 cm?
negative x direction
x (cm) 1 2 3 4 5 6 7Explanation / Answer
(a)
proton is to be released at x = 3.5 cm with initial kinetic energy 4.68 eV
moving in nagetive axis direction
Work done in bringing particles from 3.5cm to 3cm
W = F *dx = eE * dx = e (-dV/dx) * dx = 0 , (dV/dx) = 0
Work done in bringing particles from 3cm to 1cm
Assuming Vs is the topmost point in the graph image
W = F *dx = eE * dx = e (-dV/dx) * dx = e * 4.8 * 2 = 9.6 eV , (dV/dx) = - 9.6/2 V/cm
Therefore proton does not escape from the plotted region
Turining point is when its initial energy is equal to work done
4.68 eV = e (-dV/dx) * dx = 4.8eV * dx =====> dx = 0.975 cm
Therfore turining point is x = 3 -0.975 = 2.025 cm
(b) If it is initially moving in the positive direction of the axis
Work done in bringing particles from 3.5cm to 5 cm is zero
Work done in bringing particles from 5cm to 6cm
W = F *dx = eE * dx = e (-dV/dx) * dx = e * 3.2 * 1 = 3.2 eV , (dV/dx) = -3.2/1 V/cm
Hence it escapes
Its speed at x = 6.0 cm
E = 4.68 -3.2 = 1.48 eV = 1/2 m v2 = 1.684 *104 m/s
(c) What is the magnitude F of the electric force on the proton if the proton moves just to the left of x = 3.0 cm?
F = eE = e (-dV/dx) = 4.8 eV/cm
(d)
Direction of electric force = + ( Right)
e)
F = 3.2eV/cm
f)
Firection at 5 cm is === - (left)
(its a potential well)
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