a) To determine whether the particle is a proton or an electron, we need to know the sign of the charge on the particle. Since the particle is moving from the left plate (which has a more negative voltage) to the right plate (which has a less negative voltage), we know that the particle is negatively charged. Therefore, the particle is an electron.
b) We can use the conservation of energy to determine the speed of the electron just as it gets to the second plate. At any given point, the kinetic energy of the electron is given by:
KE = (1/2)mv^2
where m is the mass of the electron and v is its speed. The potential energy of the electron is given by:
PE = qV
where q is the charge on the electron and V is the voltage difference between the two plates.
The total energy of the electron (the sum of its kinetic and potential energies) is conserved, so we can write:
KE + PE = constant
Initially, the electron has a kinetic energy of:
KE1 = (1/2)mv1^2 = (1/2)(9.11 × 10^-31 kg)(90,000 m/s)^2 = 3.88 × 10^-18 J
and a potential energy of:
PE1 = qV1 = (1.60 × 10^-19 C)(-70 V) = -1.12 × 10^-17 J
where we have used the charge of the electron as q.
As the electron moves to the right plate, it loses potential energy and gains kinetic energy. When it reaches the right plate, its potential energy is:
PE2 = qV2 = (1.60 × 10^-19 C)(-50 V) = -8.00 × 10^-18 J
Since the total energy is conserved, we can write:
KE1 + PE1 = KE2 + PE2
Substituting in the values we have calculated and solving for v2, we get:
(1/2)(9.11 × 10^-31 kg)(90,000 m/s)^2 + (1.60 × 10^-19 C)(-70 V) = (1/2)(9.11 × 10^-31 kg)v2^2 + (1.60 × 10^-19 C)(-50 V)
Simplifying and solving for v2, we get:
v2 = 58,539 m/s
Therefore, the speed of the electron just as it gets to the second plate is 58,539 m/s.