An electron is accelerated horizontally from rest in atelevision picture tube by a potential difference of 5710 V. Itthen passes between two horizontal plates 6.66 cm long and 1.29 cmapart that have a potential difference of 256 V (see figure below). At what angle q will the electron betraveling after it passes between the plates?

Question

An electron is accelerated horizontally from rest in atelevision picture tube by a potential difference of 5710 V. Itthen passes between two horizontal plates 6.66 cm long and 1.29 cmapart that have a potential difference of 256 V (see figure below).  At what angle q will the electron betraveling after it passes between the plates?

Alara Mccarthy · Accepted Answer

Charge q when subjected to a potential difference of V willacquire an energy Vq Joles so electron energy is  Vq={12}mv2  v=2Vqm  =2×5710×1.6×10−199.1×10−31  =44.8097×106ms  At the time of entering the plates the electron has vertical velocity component zero and horizontal component of velocity given above  vx=44.8097×106 ms  vy=0  acceleration  ay=Fm=QEm  E is the electric field between plates Q is charge and m is mass of the particle. Also, if it is time of flight between the plates  y=12ayt2  and  L=vxt  where y is vertical distance travelled andL is plate length eliminating it from the above two equations and substitutingacceleration value  y=12QEm(Lvx)2  =12×1.6×10−199.1×10−31×2561.29100×(6.6610044.8097×106)2  =0.06m  tan⁡(θ)=0.066.66100=0.9009

An electron is accelerated horizontally from rest in atelevision picture tube by a potential

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