In Figure below, an electron accelerated from rest throughpotential difference V

Question

In Figure below, an electron accelerated from rest throughpotential difference V1=1.34 kV enters the gapbetween two parallel plates having separation d = 26.6 mmand potential difference V2= 99.9 V. The lowerplate is at the lower potential. Neglect fringing and assume thatthe electron's velocity vector is perpendicular to the electricfield vector between the plates. In unit-vector notation, whatuniform magnetic field allows the electron to travel in a straightline in the gap?

Chapter 28, Problem 7

Explanation / Answer

Given that Charge of the electron (q1) =1.6*10-19C Mass of the electron (m) =9.11*10-31kg The initital potential difference (V1) = 1.34kV                                                        = 1.34 x103                                                     = 1340V The potential difference after the separation(V2) = 99.9 V The seperation between the two parallel paltesis (d) = 26.6 mm                                                                                   = 26.6 x10-3 m                                                                               = 0.0266m The electric field between the plates is :                         E = V2 / d                             =  99.9 / 0.0266m                             =  3755.6 V/ m Pointing from top to bottem. Now the directiopn of the Magnetic field is perpendicular tothe plane of paper. Here we know that B = E / v      Here  v = velocity itis given by            (1/2)mv2 = q1V1                      v2 = 2(1.60 x10-19* 1340V ) /( 9.11 x 10-31 )                        v  =21.695 x 106          B = E /V               = 3755.6 / 21.695 x106                  = 173.109*10-6T                =0.173mT If the electron is more undeflected then magnetic field isgiven by :           B = 0.173mT( - k )                                                     = 1340V The potential difference after the separation(V2) = 99.9 V The seperation between the two parallel paltesis (d) = 26.6 mm                                                                                   = 26.6 x10-3 m                                                                               = 0.0266m The electric field between the plates is :                         E = V2 / d                             =  99.9 / 0.0266m                             =  3755.6 V/ m Pointing from top to bottem. Now the directiopn of the Magnetic field is perpendicular tothe plane of paper. Here we know that B = E / v      Here  v = velocity itis given by            (1/2)mv2 = q1V1                      v2 = 2(1.60 x10-19* 1340V ) /( 9.11 x 10-31 )                        v  =21.695 x 106          B = E /V               = 3755.6 / 21.695 x106                  = 173.109*10-6T                =0.173mT If the electron is more undeflected then magnetic field isgiven by :           B = 0.173mT( - k )

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