Two parallel conducting grids AA and BB are set at different electric potentials

Question

Two parallel conducting grids AA and BB are set at different electric potentials (Fig.2), the first at V1 = 100 V and the second at V2 = 400 V. The grids delimit three regions of the plane: to the left of AA the potential is constant and equal to V1, to the right of BB the potential is constant and equal to V2, in the middle an electric field E exists.

An electron beam enters the system from left (beam 'e') with an angle ?1 = 30° with respect to the normal to the grids, and kinetic energy ?1 = 102 eV. When electrons enter into the region at potential V2 they have an angle ?2 with respect to the normal.

a) Derive the relation between ?1 and ?2, expressing the ratio sin ?1 / sin ?2 as afunction of the other system parameters (V1, V2, ?1). Calculate the exit angle ?2 (in degrees) of electrons, and their velocity at the exit v2 (in m/s).

Explanation / Answer

Let m be the mass of electron

Initial velocity is v1 at the left edge.

The horizontal component is v1Cos1 = 0.866 v1 and vertical component of velocity is v1Sin 1 = 0.5 v1

If total initial Kinetic energy is 0.5m v12 = 1

then Initial Kinetic energy due to the horizontal component = 0.5 m (v1Cos1)2 = 1 Cos2 1

Note that since the potential different is horizontal there will be no effect on the vertical component of velocity, but the horizontal component will increase.

Increase in the Kinetic energy due to potential raise = Potential difference x charge = (V2 - V1 )x e

So the final kinetic energy due to horizontal component of velocity v2 = 1 Cos2 1 + (V2 - V1 )e

But horizontal component of velocity in terms of v2 = v2Cos 2

Therefore, if m is the mass of the particle in the beam (here electron),

0.5mx v12 = 1 (initial Kinetic energy)   eqn 1

0.5mx (v2Cos2 )2 = 1 Cos2 1 + (V2 - V1 )e eqn2

But since the vertical component of velocity does not change, we can write

Intial vertical component of velocity = final vertical component of velocity

v1Sin  1 = v2 Sin  2 = > v2 = v1 ( Sin  1 / Sin  2 ) eqn 3

Substituting, eqn 3 in eqn 2, we get,

0.5mx (v1 ( Sin  1 / Sin  2 )Cos2 )2 = 1 Cos2 1 + (V2 - V1 )e

0.5mx v12 ( ( Sin  1 / Sin  2 )Cos2 )2 = 1 Cos2 1 + (V2 - V1 )e   

But from eqn 1 0.5 m v12 = 1 , substituting this, we get,

1 ( ( Sin  1 / Sin  2 )Cos2 )2 = 1 Cos2 1 + (V2 - V1 )e

1 ( Sin  1 / Sin  2 ) 2 ( 1 - Sin22 ) = 1 (1 - Sin2 1 ) + (V2 - V1 )e

( Sin  1 / Sin  2 ) 2 - Sin2 1 = 1 - Sin2 1 + (V2 - V1 )e / 1

( Sin  1 / Sin  2 ) 2 =    1 + (V2 - V1 )e / 1

Sin  1 / Sin  2 = ( 1 + (V2 - V1 )e / 1 )1/2

Substituting 1 = 30 degrees, V2=400 , V1 = 100 , and 1 = 102 e in the above result,

Sin 30 / Sin 2 = ( 1 + 300e / 102e ) 1/2 = (1 + 300/102 )1/2 = 1.985239651

Sin 2 = Sin 30 / 1.985239651 = 0.251858761

2 = Sin-1   0.251858761 = 14.58 degrees

0.5 m v12 = 1 = 102 eV = 102 x 1.60218 x 10-19 Joules

0.5 x 9.10938356 × 10-31 x v12   = 102 x 1.60218 x 10-19

=> v1 = 5989992 m/s

v2 = v1 ( Sin  1 / Sin  2 ) from eqn 3

v2 = 5989992 ( Sin 30 / Sin 14.58 )

v2 = 11891569.63 m/s

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