Question

An electron is fired at a speed of v_0 = 5.6\times10^6 m/s and at an angle of \theta_0=–45^\circ between two parallel conductingplates that are D=2.0

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asked 2021-05-12

An electron is fired at a speed of \(\displaystyle{v}_{{0}}={5.6}\times{10}^{{6}}\) m/s and at an angle of \(\displaystyle\theta_{{0}}=–{45}^{\circ}\) between two parallel conductingplates that are D=2.0 mm apart, as in Figure. Ifthe potential difference between the plates is \(\displaystyle\triangle{V}={100}\ {V}\), determine (a) how close d the electron will get to the bottom plate and (b) where the electron will strike the top plate.
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Answers (1)

2021-05-14

Initial speed of the electron is \(\displaystyle{v}_{{0}}={5.6}\times{10}^{{6}}\) m/s and \(\displaystyle\theta_{{0}}=-{45}^{\circ}\)
The distance betweeen two parallel conducting plates is \(\displaystyle{D}={2.0}\ {m}{m}={2.0}\cdot{10}^{{-{3}}}\ {m}\)
The potential difference between the plates is \(\displaystyle\triangle{V}={100}\ {V}\)
Now the direction of the electric field is along the Y-direction.
Now the electric field strength can be calculated by using the formula
\(\displaystyle{E}_{{y}}={\frac{{\triangle{V}}}{{{D}}}}={\frac{{{100}}}{{{2}\times{10}^{{-{3}}}}}}={5000}\ \frac{{N}}{{m}}\)
The acceleration of the electron is calculated as
\(\displaystyle{a}_{{y}}={\frac{{{F}_{{y}}}}{{{m}_{{e}}}}}={\frac{{{E}_{{y}}{q}}}{{{m}_{{e}}}}}\)
\(\displaystyle={\frac{{{\left({5000}\right)}{\left({1.6}\times{10}^{{-{19}}}\right)}}}{{{9.1}\times{10}^{{-{31}}}}}}={8.78}\times{10}^{{{15}}}\ \frac{{m}}{{s}^{{2}}}\)
(a) The initial velocity v0 can beresolved into two components.
Horizontal component is \(\displaystyle{v}_{{0}}{\cos{\theta}}\)
Vertical component is \(\displaystyle{v}_{{0}}{\sin{\theta}}\)
At the bottom plate the final velocity of the electron is zero.
That means \(\displaystyle{v}_{{y}}={0}\)
Now the vertical displacement of the electron is calculated byusing the formula
\(\displaystyle{{v}_{{y}}^{{2}}}={{v}_{{{0}{y}}}^{{2}}}+{2}{a}_{{y}}{\left(\triangle{y}\right)}\)
\(\displaystyle\triangle{y}={\frac{{-{{v}_{{{0}{y}}}^{{2}}}}}{{{2}{a}_{{y}}}}}\)
\(\displaystyle={\frac{{-{\left({v}_{{0}}{\sin{\theta}}\right)}^{{2}}}}{{{2}{a}_{{y}}}}}\)
\(\displaystyle={\frac{{-{\left[{\left(-{5.6}\times{10}^{{6}}\ \frac{{m}}{{s}}\right)}{\sin{{45}}}\right]}^{{2}}}}{{{2}{\left({8.78}\times{10}^{{{15}}}\right)}}}}\)
\(\displaystyle=-{0.89}\) mm
Now the minimum distance above the bottom plate is then
\(\displaystyle{d}={\frac{{{D}}}{{{2}}}}+\triangle{y}={\frac{{{2.0}\times{10}^{{-{3}}}}}{{{2}}}}+{\left(-{0.89}{m}{m}\right)}={0.11}{m}{m}\)
b) The time for electron to travel from the point O to the upperplate is ccalculated as \(\displaystyle\triangle{y}={v}_{{{0}{y}}}{t}+{\frac{{{1}}}{{{2}}}}{a}_{{y}}{t}^{{2}}\)
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