A proton moves from location A to location B in a region of uniform electric field, as shown in Figure 16.5. (a) If the magnitude of the electric field inside the capacitor in Figure 16.5 is 3500 N/C, and the distance between location A and location B is 3 mm, what is the change in electric potential energy of the system (proton + plates) during this process? (b) What is the change in the kinetic energy of the proton during this process? (c) If the proton is initially at rest, what is its speed when it reaches location B? (d) How do the answers to (a) and (b) change if the proton is replaced by an electron?

When a charge moves between two electrically charged plates then the change in the amount of potential energy of the plate and charge system relies upon the electric field between the plates and their distance from each other.

If the distance between charged plates is increased then the potential energy of the system also increases.

The magnitude of the electric field inside the capacitor is,$\mathrm{E}=3500\mathrm{N}/\mathrm{C}$.

The distance between location A and location B is

$d=3mm\times \frac{1m}{1000mm}=0.003m$

The amount of the charge of a proton is given by,

$q_p=1.6x{10}^{-19}\mathrm{C}$

The formula for thechange in electric potential energy of the system is given by,

$$\begin{gathered} ∆\mathrm{U}=-{\mathrm{q}}_{\mathrm{p}}\times \mathrm{E}\times \mathrm{d} \\ ∆\mathrm{U}=-\left(1.6\times {10}^{-19}\mathrm{C}\right)3500\mathrm{N}/\mathrm{C}\times 0.003\mathrm{m}\times \frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}} \\ ∆\mathrm{U}=-16.8\times {10}^{-19}\mathrm{J} \end{gathered}$$

Hence, the change in electric potential energy of the system is$-16.8\times {10}^{-19}\mathrm{J}$.

The formula for thechange in the kinetic energy of the proton is given by,

$$\begin{gathered} ∆\mathrm{K}+∆\mathrm{U}=0 \\ ∆\mathrm{K}=-∆\mathrm{U} \\ ∆\mathrm{K}=-\left(-16.8\times {10}^{-19}\mathrm{J}\right) \\ ∆\mathrm{K}=-16.8\times {10}^{-19}\mathrm{J} \end{gathered}$$

Hence, the change in the kinetic energy of the proton is $16.8\times 10\mathrm{J}$.

The formula for the change in the kinetic energy of the proton is given by,

$∆K=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

Proton is initially at rest, so $u=0$and its mass is, $\mathrm{m}=1.67\times {10}^{-27}\mathrm{kg}$.

$$\begin{gathered} 16.8\times {10}^{-19}J=\frac{1}{2}\times 1.67\times {10}^{-27}kg\times v^2-0 \\ {v}^2=\frac{16.8\times {10}^{-19}J}{\frac{1}{2}\times 1.67\times {10}^{-27}kg}\times \frac{1m^2/s^2}{1J/kg} \\ {v}^2=20.119\times {10}^8{m}^2/s^2 \\ v=\sqrt{20.119\times {10}^8{m}^2/s^2} \\ v=4.485\times {10}^{4}m/s \end{gathered}$$

Hence, the final speed of the proton when it reaches location B is $4.485\times {10}^{4}m/s$.

If the proton is replaced by electron, then the charge of an electron will be,

$q_e=-1.6x{10}^{-19}C$

Then,thechange in electric potential energy of the system is given by,

$$\begin{gathered} ∆\mathrm{U}=-{\mathrm{q}}_e\times \mathrm{E}\times \mathrm{d} \\ ∆\mathrm{U}=-\left(-1.6\times {10}^{-19}\mathrm{C}\right)3500\mathrm{N}/\mathrm{C}\times 0.003\mathrm{m}\times \frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}} \\ ∆\mathrm{U}=16.8\times {10}^{-19}\mathrm{J} \end{gathered}$$

And, the change in the kinetic energy of the proton is given by,

$$\begin{gathered} ∆\mathrm{K}+∆\mathrm{U}=0 \\ ∆\mathrm{K}=∆\mathrm{U} \\ ∆\mathrm{K}=-\left(16.8\times {10}^{-19}\mathrm{J}\right) \\ ∆\mathrm{K}=-16.8\times {10}^{-19}\mathrm{J} \end{gathered}$$

Hence, the answers of (a) and (b) if the proton is replaced by an electron changes to $16.8\times {10}^{-19}\mathrm{J}$and $-16.8\times {10}^{-19}\mathrm{J}$respectively.