An electron is fired at a speed of v0=5.6×106 m/s and at an angle of θ0=–45∘ between two parallel conductingplates that are D=2.0 mm apart, as in Figure. Ifthe potential difference between the plates is △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.

Question

An electron is fired at a speed of v0=5.6×106 m/s and at an angle of θ0=–45∘ between two parallel conductingplates that are D=2.0 mm apart, as in Figure. Ifthe potential difference between the plates is △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.

Aubree Mcintyre · Accepted Answer

Initial speed of the electron is v0=5.6×106 m/s and θ0=−45∘ The distance betweeen two parallel conducting plates is D=2.0 mm=2.0⋅10−3 m The potential difference between the plates is △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 Ey=△VD=1002×10−3=5000 Nm The acceleration of the electron is calculated as ay=Fyme=Eyqme =(5000)(1.6×10−19)9.1×10−31=8.78×1015 ms2 (a) The initial velocity v0 can beresolved into two components. Horizontal component is v0cos⁡θ Vertical component is v0sin⁡θ At the bottom plate the final velocity of the electron is zero. That means vy=0 Now the vertical displacement of the electron is calculated byusing the formula vy2=v0y2+2ay(△y) △y=−v0y22ay =−(v0sin⁡θ)22ay =−[(−5.6×106 ms)sin⁡45]22(8.78×1015) =−0.89 mm Now the minimum distance above the bottom plate is then d=D2+△y=2.0×10−32+(−0.89mm)=0.11mm b) The time for electron to travel from the point O to the upperplate is ccalculated as △y=v0yt+12ayt2

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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