A pair of charged conducting plates produces a uniform field of E o = 11,751 N/C

Question

A pair of charged conducting plates produces a uniform field of Eo = 11,751 N/C directed to the right, between the plates. The separation of the plates is L = 12 mm. In Figure, an electron (e = - 1.6 x 10-19 C; m = 9.1 x 10-31 kg) is projected from plate A, directly toward plate B, with an initial velocity ofvo = 2

A pair of charged conducting plates produces a uniform field of Eo = 11,751 N/C directed to the right, between the plates. The separation of the plates is L = 12 mm. In Figure, an electron (e = - 1.6 times 10-19 C; m = 9.1 times 10-31 kg) is projected from plate A, directly toward plate B, with an initial velocity of vo = 2 times 107 m/s. The velocity of the electron (expressed in general form as a whole number) as it strikes plate B is:

Explanation / Answer

acceleration of electron = qE / m = 1.6 * 10^-19 * 11751 / 9.1*10^-31 = 2066109890109890.1098 m/s^2

velocity as it strikes plate B is given by

v^2 - u^2 = 2*a*S

Vx^2 - 0^2 = 2* 2066109890109890.1098*0.012

so, Vx = sqrt(2066109890109890.1098*2*0.012) = 7041777.9972 m/s

so velocity = sqrt ( vx^2 + Vy^2) = sqrt( 7041777.9972^2 + (2*10^7)^2) = 21203458.14629 m/s is the answer

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