A problem of practical interest is to make a beam of electrons turn a 90 corner.

Question

A problem of practical interest is to make a beam of electrons turn a 90 corner. This can be done with the parallel-plate capacitor shown in the figure (Figure 1) . An electron with kinetic energy 2.0×1017 J enters through a small hole in the bottom plate of the capacitor.

Should the bottom plate be charged positive or negative relative to the top plate if you want the electron to turn to the right?

What strength electric field is needed if the electron is to emerge from an exit hole 1.0 cm away from the entrance hole, traveling at right angles to its original direction?

Explanation / Answer

force on the electron due to the electric field should be in downward direction

as force on a charge=electric charge*electric field

and charge of electron is negative, the direction of electric field be opposite to that of the direction of force

hence electric field should be from bottom plate to top plate.

so bottom plate is needed to be positively charged.

part B:

mass of electron=m=9.1*10^(-31) kg

let initial speed of electron be v.

then 0.5*m*v^2=2*10^(-17)

==>v=6.63*10^6 m/s

coordinate system:

let the x axis be along the bottom plate

y axis be perpendicular to the bottom plate

force is along y axis and in downward direction

acceleration=a=q*E/m

acceleration along x axis=0

so velocity along x axis will remain constant=v*cos(45)=4.6881*10^6 m/s

initial velocity along y axis=v*sin(45)=4.6881*10^6 m/s

distance to be covered along x axis=1 cm=0.01 m

then time taken=distance/velocity=0.01/(4.6881*10^6)=2.1331 ns

distance covered along y axis during that time=0

==>initial speed*time+0.5*acceleration*time^2=0

==>4.6881*10^6*2.1331*10^(-9)-0.5*a*(2.1331*10^(-9))^2=0

==>a=4.3956*10^15 m/s^2

==>q*E/m=4.3956*10^15

==>E=4.3956*10^15*9.1*10^(-31)/(1.6*10^(-19))=2.5*10^4 N/C

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