The two parallel plates in FIGURE P$23.53$are $2.0\mathrm{cm}$apart and the electric field strength between them is $1.0\times {10}^4\mathrm{N}/\mathrm{C}$. An electron is launched at a ${45}^{\circ }$ angle from the positive plate. What is the maximum initial speed $v_0$ the electron can have without hitting the negative plate?

$\overrightarrow{E}=1.0\times {10}^4\mathrm{N}/\mathrm{C}\text{- magnitude of the electric field strength}$

$m_e=9.11\times {10}^{-31}\mathrm{kg}\text{- mass of electron}$

$q=1.6\times {10}^{-19}\mathrm{C}-\text{electron charge}$

$d=x=2\mathrm{cm}=2.0\times {10}^{-2}\mathrm{m}$$\text{- distance between + and - plate}$

We must make a decision$v_{i_{max}}$ the electron's

The path coefficient is used:

$\overrightarrow{a}=\frac{q}{m}\overrightarrow{E}$

We see from it in a magnetic charge, a charge will face an electrostatic force, prompting the particle (electron) to accelerated

$\overrightarrow{a}=\frac{1.6\times {10}^{-19}\mathrm{C}}{9.11\times {10}^{-31}\mathrm{kg}}\left(1.0\times {10}^4\mathrm{N}/\mathrm{C}\right)$

$=1.8\times {10}^{15}\mathrm{m}/{\mathrm{s}}^2$

The momentum is described by a simple interaction:

$v_f^2-v_i^2=2ax$

The mobility of such an entity with constant speed is indicated by it connection.

The average speed inside this direction $y$indicated as:

$V_{iy}=v_0\sin \left({45}^{\circ }\right)$

$V_y^{20}-V_{iy}^2=2ax$

$V_{iy}^2=2ad$

$v_o^2{\sin }^2\left({45}^{\circ }\right)=2ad$

$v_0=\frac{\sqrt{2ad}}{\sin \left({45}^{\circ }\right)}$

Then should establish the actual peak value start pace of the ion by inputting all given data.

$v_0=\frac{\sqrt{2\left(1.8\times {10}^{15}\mathrm{m}/{\mathrm{s}}^2\right)\left(2.0\times {10}^{-2}\mathrm{m}\right)}}{\sin \left({45}^{\circ }\right)}$

$=1.2\times {10}^7\mathrm{m}/\mathrm{s}$