Question

field. Because the electrons are nearly at rest initially, one can calculate the velocity of the electrons after this acceleration from conservation of energy. ? K.E. = -? P.E. (1) Where K.E. is the kinetic energy and P.E. is the potential energy. This gives: ½ m_e v_z^2 = eV_z . (2) V_z is the potential difference or voltage difference that the electron has “dropped” through; e is the charge of the electron = 1.6 x 10?¹? Coulomb; and m_e is the mass of an electron = 9.11 x 10?³¹ kg. An intuitive way of understanding this, is to think of the electron as a marble rolling down the voltage (or potential energy) curve shown at the bottom of fig. 1. After this initial acceleration, an electrostatic lens focuses the electrons so that they will hit the screen in a small region. Next the electrons pass between two pairs of deflection plates; see fig. 1 and 2. Each pair of plates, is like a parallel plate capacitor and there is an electrical potential, V, between the plates. These plates are separated by a distance, d. Q i. What is the electric field, E_y, in the region between the plates? E_y(V, d) = ????? (3) Q ii. What force, F?, (magnitude and direction) does an electron experience when it is between the plates because of this electric field? F? = ????? (4) Q iii. What acceleration, a?, does the electron experience due to this force (F=ma)? a? = ????? (5) Q iv. If the electron is between the plates for a time, t, what is the y-component of the velocity after the electron exits from between the plates, v_{final, y}? v_{final, y}(a_y, t) = ????? (6) Q v. What is the relationship between the z-components of the initial and final velocities? Q vi. What time, t, is the electron between the deflection plates if the plates are a length, ?? t(v_{initial, z}, ?) = ????? (7) Q vii. What is the v_{final, y}, in terms of the plate separation, d; the plate length, ?; the deflection plate voltage drop, V; the velocity along the z-axis, v_{initial, z}; the electron’s mass, m_e; and the charge, e? v_{final, y}(d, ?, V, v_{initial, z}, m_e, e) = ????? (8) The deflection angle, ?, is given by: tan(?) = v_{final, y} / v_{final, z} (9) If L >> ? then: D/L ? tan(?) . (10) And this approximation can be improved if L is replaced by L' = L + ?/2 .

          field. Because the electrons are nearly at rest initially, one can calculate the velocity of the electrons after this acceleration from conservation of energy.
? K.E. = -? P.E. (1)
Where K.E. is the kinetic energy and P.E. is the potential energy. This gives:
½ m_e v_z^2 = eV_z . (2)
V_z is the potential difference or voltage difference that the electron has “dropped” through; e is the charge of the electron = 1.6 x 10?¹? Coulomb; and m_e is the mass of an electron = 9.11 x 10?³¹ kg. An intuitive way of understanding this, is to think of the electron as a marble rolling down the voltage (or potential energy) curve shown at the bottom of fig. 1. After this initial acceleration, an electrostatic lens focuses the electrons so that they will hit the screen in a small region.

Next the electrons pass between two pairs of deflection plates; see fig. 1 and 2. Each pair of plates, is like a parallel plate capacitor and there is an electrical potential, V, between the plates. These plates are separated by a distance, d.
Q i. What is the electric field, E_y, in the region between the plates?
E_y(V, d) = ????? (3)
Q ii. What force, F?, (magnitude and direction) does an electron experience when it is between the plates because of this electric field?
F? = ????? (4)

Q iii. What acceleration, a?, does the electron experience due to this force (F=ma)?
a? = ????? (5)
Q iv. If the electron is between the plates for a time, t, what is the y-component of the velocity after the electron exits from between the plates,
v_{final, y}?
v_{final, y}(a_y, t) = ????? (6)
Q v. What is the relationship between the z-components of the initial and final velocities?
Q vi. What time, t, is the electron between the deflection plates if the plates are a length, ??
t(v_{initial, z}, ?) = ????? (7)
Q vii. What is the v_{final, y}, in terms of the plate separation, d; the plate length, ?; the deflection plate voltage drop, V; the velocity along the z-axis, v_{initial, z}; the electron’s mass, m_e; and the charge, e?
v_{final, y}(d, ?, V, v_{initial, z}, m_e, e) = ????? (8)

The deflection angle, ?, is given by:
tan(?) = v_{final, y} / v_{final, z} (9)
If L >> ? then:
D/L ? tan(?) . (10)
And this approximation can be improved if L is replaced by L' = L + ?/2 .

field. Because the electrons are nearly at rest initially, one can calculate the velocity of the electrons after this acceleration from conservation of energy.
? K.E. = -? P.E. (1)
Where K.E. is the kinetic energy and P.E. is the potential energy. This gives:
½ me vz^2 = eVz . (2)
Vz is the potential difference or voltage difference that the electron has “dropped” through; e is the charge of the electron = 1.6 x 10?¹? Coulomb; and me is the mass of an electron = 9.11 x 10?³¹ kg. An intuitive way of understanding this, is to think of the electron as a marble rolling down the voltage (or potential energy) curve shown at the bottom of fig. 1. After this initial acceleration, an electrostatic lens focuses the electrons so that they will hit the screen in a small region.

Next the electrons pass between two pairs of deflection plates; see fig. 1 and 2. Each pair of plates, is like a parallel plate capacitor and there is an electrical potential, V, between the plates. These plates are separated by a distance, d.
Q i. What is the electric field, Ey, in the region between the plates?
Ey(V, d) = ????? (3)
Q ii. What force, F?, (magnitude and direction) does an electron experience when it is between the plates because of this electric field?
F? = ????? (4)

Q iii. What acceleration, a?, does the electron experience due to this force (F=ma)?
a? = ????? (5)
Q iv. If the electron is between the plates for a time, t, what is the y-component of the velocity after the electron exits from between the plates,
vfinal, y?
vfinal, y(ay, t) = ????? (6)
Q v. What is the relationship between the z-components of the initial and final velocities?
Q vi. What time, t, is the electron between the deflection plates if the plates are a length, ??
t(vinitial, z, ?) = ????? (7)
Q vii. What is the vfinal, y, in terms of the plate separation, d; the plate length, ?; the deflection plate voltage drop, V; the velocity along the z-axis, vinitial, z; the electron’s mass, me; and the charge, e?
vfinal, y(d, ?, V, vinitial, z, me, e) = ????? (8)

The deflection angle, ?, is given by:
tan(?) = vfinal, y / vfinal, z (9)
If L >> ? then:
D/L ? tan(?) . (10)
And this approximation can be improved if L is replaced by L' = L + ?/2 .

Added by Kerry C.

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