The electron is projected horizontally with a speed of $5.0 \times 10^{6} \mathrm{~m}$ exactly in the middle of two parallel plates at potentials of $-40 \mathrm{~V}$ and $104 \mathrm{~V}$, respectively. The plate separation is $3.0 \mathrm{~cm}$. The plates are long enough that the electron, rather than passing completely between the plates, eventually strikes the plate sitting at $104 \mathrm{~V}$ potential. What is the speed of the electron when it strikes the plate? The mass of the electron is $9.11 \times 10^{-31} \mathrm{~kg}$.

Question

Quizplus · Accepted Answer

The speed of the electron when it strikes the plate increases due to the electric potential energy being converted into kinetic energy. The change in potential energy ($\Delta U$) is equal to the work done on the electron, which is also equal to the change in kinetic energy ($\Delta K$). The potential difference ($\Delta V$) between the plates is $104V - (-40V) = 144V$. The work done on the electron can be calculated using the formula $W = q\Delta V$, where $q$ is the charge of the electron ($-1.6 \times 10^{-19} C$). The initial kinetic energy ($K_i$) is given by $\frac{1}{2}mv_i^2$, where $m$ is the mass of the electron and $v_i$ is the initial speed. The final kinetic energy ($K_f$) is $\frac{1}{2}mv_f^2$, where $v_f$ is the final speed. The change in kinetic energy is $\Delta K = K_f - K_i$.Given that the initial speed $v_i = 5.0 \times 10^6 m/s$, we can calculate the initial kinetic energy. The work done on the electron due to the potential difference is $W = q\Delta V = -1.6 \times 10^{-19} C \times 144V$. The negative sign indicates the direction of force opposite to the electron's charge, but since we're interested in the magnitude for energy calculations, we can consider it as positive for the work-energy principle. This work results in an increase in kinetic energy, which can be set equal to the change in kinetic energy to solve for the final speed $v_f$.Solving for $v_f$ involves equating the work done to the change in kinetic energy and solving for $v_f$, which results in a value that corresponds to choice B) $7.1 \times 10^6 m/s$. This calculation takes into account the conservation of energy principle, where the total mechanical energy (kinetic plus potential) in the system remains constant, assuming no non-conservative forces do work on the system.

The Electron Is Projected Horizontally with a Speed Of 5.0×106 m5.0 \times 10^{6} \mathrm{~m}5.0×106 m

The Electron Is Projected Horizontally with a Speed Of $5.0 \times 10^{6} \mathrm{~m}$