91影视

1. Separation between the two parallel plates,
2. The potentials of the plates,${\mathrm{V}}_1=-70.0\mathrm{volt}$and${\mathrm{V}}_2=-50.0\mathrm{volt}$
3. Initial speed of the particle,${\mathrm{v}}_{\mathrm{o}}=\left(90.0\mathrm{km}/\mathrm{s}\right)=90.0脳{10}^3\mathrm{m}/\mathrm{s}$
4. Charge of the electron or proton,$\mathrm{q}=1.6脳{10}^{-19}\mathrm{C}$
5. Mass of the electron or proton,$\mathrm{m}=1.67脳{10}^{-27}\mathrm{kg}$

The nature of the particle can be determined as follows. A charged particle is moving in between the parallel plates separated by a distance of d as shown in the following figure. The particle is released with some initial velocity slows down due to the potential difference between the plates and reached the left plate with a final velocity that is less than the initial velocity. The following figure shows a charged particle moving between two parallel plates.

The right plate is at a higher potential than the left plate, so the electric field is directed towards the left as shown in the above figure.

According to the law of conservation of energy, the initial potential energy, and kinetic energy should be equal to the final kinetic and potential energies.

Formulae:

The kinetic energy of the particle, $\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2$ 鈥�(颈)

The potential energy of the particle, data-custom-editor="chemistry" $\mathrm{U}=\mathrm{qV}$ 鈥�(颈颈)

Since the speed of the particle decreases as it moves from the left side to the right side, then the force should be in the opposite direction to the motion of the particle. Hence, both electric field and electric force are parallel to each and the particle decelerates as it moves towards the second plate, it should be a positive particle.

Therefore, the particle is a proton.

Using equations (i) and (ii) in the conservation of energy of the system, the speed of the particle can be determined as follows: (Here mass of the particle is m , initial speed of the particle is v_o, charge of the particle is q , potential of left plate is V₁, potential of the right particle is V₂, and final speed of the particle is v)

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{o}}+{\mathrm{U}}_{\mathrm{o}}& =& {\mathrm{K}}_{\mathrm{f}}+{\mathrm{U}}_{\mathrm{f}}\\ \frac{1}{2}{\mathrm{mv}}_{\mathrm{o}}^2+{\mathrm{qV}}_1& =& \frac{1}{2}{\mathrm{mv}}^2+{\mathrm{qV}}_2\\ \frac{1}{2}{\mathrm{mv}}_{\mathrm{o}}^2+{\mathrm{qV}}_1-{\mathrm{qV}}_2& =& \frac{1}{2}{\mathrm{mv}}^2\\ \frac{1}{2}{\mathrm{mv}}_{\mathrm{o}}^2+\mathrm{q}\left({\mathrm{V}}_1-{\mathrm{V}}_2\right)& =& \frac{1}{2}{\mathrm{mv}}^2\\ {\mathrm{v}}^2& =& {\mathrm{v}}_{\mathrm{o}}^2+\frac{2\mathrm{q}\left({\mathrm{V}}_1-{\mathrm{V}}_2\right)}{\mathrm{m}}\\ \mathrm{v}& =& \sqrt{{\mathrm{v}}_{\mathrm{o}}^2+\frac{2\mathrm{q}\left({\mathrm{V}}_1-{\mathrm{V}}_2\right)}{\mathrm{m}}}\\ \mathrm{v}& =& \sqrt{{\left(90.0脳{10}^3\mathrm{m}/\mathrm{s}\right)}^2+\frac{2\left(1.6脳{10}^{-19}\mathrm{C}\right)\left(-70.0\mathrm{V}-\left(-50.0\mathrm{V}\right)\right)}{1.67脳{10}^{-27}\mathrm{kg}}}\\ & =& 65.33脳{10}^3\mathrm{m}/\mathrm{s}\\ & =& 65.3\mathrm{km}/\mathrm{s}\end{array}$

Therefore, the speed of the particle as it reached plate 2 is 65.33 km/s.