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 \(\displaystyle\triangle{y}={v}_{{{0}{y}}}{t}+{\frac{{{1}}}{{{2}}}}{a}_{{y}}{t}^{{2}}\)

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 \(\displaystyle\triangle{y}={v}_{{{0}{y}}}{t}+{\frac{{{1}}}{{{2}}}}{a}_{{y}}{t}^{{2}}\)