In a given CRT, electrons are accelerated horizontally by 8.5 k V . They then pa

Question

                    In a given CRT, electrons are accelerated horizontally by 8.5kV . They then pass through a uniform electric field E for a distance of 3.3cm , which deflects them upward so they travel 22cm to the top of the screen, 12cm above the center.                 

                    Estimate the value of E. Neglect the upward distance that electron                     moves while in the electric field.                 

                                     Please SHOW WORK!

Explanation / Answer

potential difference thorugh which the electron travels = 8500 V

so.. change in energy of electron = 8500 eV = 8500 * 1.6 * 10^-19 = 1.36 * 10^-15 J

so.. for electron .. after acceleration

kinetic energy = 1.36 * 10^-15

so.. 0.5 * mass_electron * v^2 = 1.36 * 10^-15

so.. 0.5 * 9.1 * 10^-31 * v^2 = 1.36 * 10^-15

so... v = 54671848.23 m/sec

now for deflection ...

along horizontal ..

distacne = 3.3 cm= 0.033 m

v = 54671848.23 m/sec

so.. time for which electric field acts = t = 0.033/54671848.23 = 6.0360132 * 10^-10 secs

so.. t = 6.0360132 * 10^-10 secs

y = 22 cm = 0.22 m

let acceleraion = a

so.. y = ut + 0.5*a*t^2

u = inital velocity in y direction = 0

so.. 0.22 = 0 + 0.5*a*(6.0360132 * 10^-10 )^2

so.. a = 1.2076812 * 10^ 18 m/sec2

so... force = mas of electron * a = 9.1 * 10^-31 * 1.20768 * 10^18 = 1.09899 * 10^-12 N

so.. electric field = force / charge of electron = 1.09899 * 10^-12 / ( 1.6 * 10^-19 ) = 6.86869 * 10^6 N/C

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