Question

Suppose electrons enter the uniform electric field midwaybetween two plates, moving at an upward 45 degree angle as shown.What maximum speed can the electrons have if they are to avoidstriking the upper plate. Ignore fringing of the field. Upper plate is negatively charged bottom plate is positivelycharged length of plate is 6cm and distance from top of top plateto bottom of bottom plate is 1cm.
\(\displaystyle{E}={5.0}\times{10}^{{3}}\frac{{N}}{{C}}\)

Answer 1

The vertical acceleration
\(\displaystyle{a}_{{y}}={\frac{{-{q}{E}}}{{{m}}}}={\frac{{-{\left({1.6}\cdot{10}^{{-{19}}}\right)}{\left({5}\cdot{10}^{{3}}\right)}}}{{{9.11}\cdot{10}^{{-{31}}}{k}{g}}}}\)
\(\displaystyle=-{8.78}\cdot{10}^{{{44}}}\frac{{m}}{{\sec{}}}\)
The vertical velocity is
\(\displaystyle{{V}_{{y}}^{{2}}}={{V}_{{{o}{y}}}^{{2}}}+{2}{a}_{{y}}{h}\)
\(\displaystyle{V}_{{o}}{\sin{{45}}}={0}+\sqrt{{-{2}{a}_{{y}}{h}}}\)
\(\displaystyle{V}_{{o}}={\frac{{{\left[-{\left(-{8.78}\cdot{10}^{{{14}}}\frac{{m}}{{s}^{{2}}}\right)}{\left({0.010}{m}\right)}\right]}^{{\frac{{1}}{{2}}}}}}{{{\sin{{45}}}}}}\)
\(\displaystyle={4.2}\cdot{10}^{{6}}\frac{{m}}{{s}}\)