Answer to Question #79490 in Atomic and Nuclear Physics for himanshu

Question #79490
A pair of charged conducting plates produces a uniform field of Eo = 10,859 N/C directed to the right, between the plates. The separation of the plates is L = 37 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 of Vo = 2.1×107 m/s. The velocity of the electron (expressed in general form as a whole number) as it strikes plate B is what?
1
Expert's answer
2018-08-02T10:25:08-0400
Force on charge q in an electric field E is F =qE.
So force on electron
F= -1.6∙〖10〗^(-19) 10859= -1.737∙〖10〗^(-15) N
The field is positive (to the right) so the force on the electron is negative (to the left).
Acceleration of electron
a= (-1.737∙〖10〗^(-15))/(9.1∙〖10〗^(-31) ) = -1.909∙〖10〗^15 m/s^2
v_f ² = v_o ² +2as
The velocity of the electron as it strikes plate B is
v_f=√((2.1∙〖10〗^7 )^2+2(-1.909∙〖10〗^15)(0.037))=1.7∙〖10〗^7 m/s.

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