Proton in a well.Figure 24-59shows electric potential Valong an xaxis.The scale of the vertical axis is set by V_s=10.0 V. A proton is to be released at x=3.5 cmwith initial kinetic energy 4.00 eV. (a) If it is initially moving in the negativedirection of the axis, does it reach a turning point (if so, what is the x-coordinate of that point) or does it escape from the plottedregion (if so, what is its speed at x=0)? (b) If it is initially movingin the positive direction of the axis, does it reach a turning point (ifso, what is the xcoordinate of that point) or does it escape from theplotted region (if so, what is its speed at x=6.0 cm)? What are the (c) magnitude Fand (d) direction (positive or negative direction ofthe xaxis) of the electric force on the proton if the proton movesjust to the left of x=3.0 cm? What is (e) Fand (f) the direction ifthe proton moves just to the right of x=5.0 cm?

1. A proton is to be released at$\mathrm{x}=3.5\mathrm{cm}$ with initial kinetic energy ${\mathrm{KE}}_{\mathrm{i}}=400\mathrm{eV}$.
2. Mass of the proton$\mathrm{m}=1.6\times {10}^{-27}\mathrm{kg}$
3. The scale of the vertical axis,${\mathrm{V}}_{\mathrm{s}}=10\mathrm{V}$

The electrostatic force F on the charge q when placed in electric field E is equal to the product of charge and magnitude of electric field.

$\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{q}\overrightarrow{\mathbf{E}}$

Using the concept of energy, we can get the turning point of the proton by studying the graph. Again, the speed can be calculated using the kinetic energy concept of the particle. Now, using the concept of electrostatic force, we can get the force acting on the proton particle. This same concept also determines the direction of the acting electrostatic force.

Formulae:

The kinetic energy of a body in motion,$\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2$ …(i)

The electrostatic force acting on a charge, $\overrightarrow{\mathrm{F}}=\mathrm{q}\overrightarrow{\mathrm{E}}$ …(ii)

When the proton is released, its energy is:

$\begin{array}{rcl}\mathrm{K}+\mathrm{U}& =& 4.0\mathrm{eV}+3.0\mathrm{eV}\\ & =& 7.0\mathrm{eV}\end{array}$

The value of potential energy is derived from the graph as follows:

Interpolating this 7.0 eV value in the region between 1.0 cm and 3.0 cm, we find the turning point at the left that is roughly at$\mathrm{x}=1.7\mathrm{cm}$as follows.

Hence, the turning point of the proton is around 1.7 cm.

There is no turning point toward the right, so the speed there is non-zero, and is given by energy conservation using equation (i) as: (K.E.= Total energy is 7eV)

$\begin{array}{rcl}\mathrm{v}& =& \sqrt{\frac{2\left(7.0\mathrm{eV}\right)}{\mathrm{m}}}\\ & =& \sqrt{\frac{2\left(7.0\mathrm{eV}\right)\left(1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)}{1.6\times {10}^{-27}\mathrm{kg}}}\\ & =& 36623.99985\mathrm{m}/\mathrm{s}\\ & \approx & 36.6\mathrm{km}/\mathrm{s}\end{array}$

Hence, the value of the speed at the turning point is 36.62 km/s.

The electric field at any point P is the (negative of the) slope of the voltage graph evaluated at P. In the region just to the left of x=3.0 cm, the field is $\overrightarrow{\mathrm{E}}=\left(+300\mathrm{V}/\mathrm{m}\right)\hat{\mathrm{i}}$

Thus, the value of the force is given using equation (ii) as: (q=charge of the proton)

$\begin{array}{rcl}\overrightarrow{\mathrm{F}}& =& \left(+300\mathrm{V}/\mathrm{m}\right)\left(1.6\times {10}^{-19}\mathrm{C}\right)\hat{\mathrm{i}}\\ & =& \left(+4.8\times {10}^{-17}\mathrm{N}\right)\hat{\mathrm{i}}\end{array}$

Hence, the magnitude of the force is $+4.8\times {10}^{-17}\mathrm{N}$.

From the calculations of part (c), we get that the force $\overrightarrow{\mathrm{F}}$ points in the +x-direction, as the electric field $\overrightarrow{\mathrm{E}}$.

In the region just to the right of x=5.0 cm, the field is$\overrightarrow{\mathrm{E}}=\left(-200\mathrm{V}/\mathrm{m}\right)\hat{\mathrm{i}}$

Thus, the value of the force is given using equation (ii) as follows:

$\begin{array}{rcl}\overrightarrow{\mathrm{F}}& =& \left(-200\mathrm{V}/\mathrm{m}\right)\left(1.6\times {10}^{-19}\mathrm{C}\right)\hat{\mathrm{i}}\\ & =& -\left(3.2\times {10}^{-17}\mathrm{N}\right)\hat{\mathrm{i}}\end{array}$

Hence, the magnitude of the force is $3.2\times {10}^{-17}\mathrm{N}$.

From the calculations of part (e), we get that the force $\overrightarrow{\mathrm{F}}$ points in the -x direction, as the electric field $\overrightarrow{\mathrm{E}}$.