The magnitude of the potential difference
between the two plates is 1.2 103 V. The distance
between the plates is 0.12 m.
(a) At what speed will an electron pass through the hole
in the positive plate?
(b) Is the electron pulled back to the positive plate once it
passes through the hole? Explain your answer.
(c) How could the apparatus be modified to accelerate
protons?
(d) Find the speed of the emerging protons in an appro-
priate apparatus.
(a) To find the speed of the electron passing through the hole in the positive plate, we can use the equation for the potential energy of an electron in an electric field:
\(PE = qV\), where
PE = potential energy
q = charge of electron = -1.6 x 10^-19 C
V = potential difference = 1.2 x 10^3 V
Setting the potential energy equal to the kinetic energy of the electron at the hole, we have:
\(\frac{1}{2}mv^2 = qV\)
Solving for v, the speed of the electron, we get:
\(v = \sqrt{\frac{2qV}{m}}\)
Substituting the values of q, V, and the mass of an electron (m = 9.11 x 10^-31 kg), we get:
\(v = \sqrt{\frac{2(-1.6 \times 10^{-19} \times 1.2 \times 10^3)}{9.11 \times 10^{-31}}}\)
\(v \approx 2.3 \times 10^6 \, m/s\)
So, the speed at which the electron passes through the hole in the positive plate is approximately 2.3 x 10^6 m/s.
(b) The electron is not pulled back to the positive plate once it passes through the hole because it gains kinetic energy from the electric field between the plates. This kinetic energy is sufficient to overcome the attraction of the positive plate and continue moving in the direction determined by its velocity.
(c) To modify the apparatus to accelerate protons, we can either adjust the potential difference between the plates or change the charge of the plates. Increasing the potential difference or making the positive plate more positively charged will accelerate protons passing through the hole.
(d) Using a similar approach as in part (a) but with the charge of a proton (q = 1.6 x 10^-19 C) and the mass of a proton (m = 1.67 x 10^-27 kg), we can find the speed of the emerging protons. Substituting the values into the equation
\(v = \sqrt{\frac{2qV}{m}}\)
\(v = \sqrt{\frac{2(1.6 \times 10^{-19} \times 1.2 \times 10^3)}{1.67 \times 10^{-27}}\)
\(v \approx 2.1 \times 10^5 \, m/s\)
Thus, the speed of the emerging protons in an appropriate apparatus is approximately 2.1 x 10^5 m/s.