(Solved): Question 3 Figure 3 shows the path of an electron after it enters horizontally the region between t ...

Question 3 Figure 3 shows the path of an electron after it enters horizontally the region between the plates of a parallel plate capacitor with a speed of \( 5.45 \times 10^{6} \mathrm{~m} / \mathrm{s} \). The conducting plates of the capacitor are horizontal and \( 2.25 \mathrm{~cm} \) long. The electric field is assumed to be uniform between the plates. Mass \( m \) of the electron \( =9.1 \times 10^{-31} \mathrm{~kg} \); Charge \( q \) of the electron \( =-1.6 \times 10^{-19} \mathrm{C} \). The electric field of the capacitor deflects the electron downwards by a distance of \( 0.618 \mathrm{~cm} \) when the electron exits the capacitor. The force of gravity and the fringing of the electric field lines at the edges of the capacitor may be neglectedr (a) Is the electric field upwards or downwards? Justify your answer. (b) On a diagram similar to Figure 3, add \( + \) and - signs to show the expected distribution of charges on the plates of the capacitor. Describe briefly in words the charge distribution. (c) Write down an equation for the force \( \vec{F} \) on a charged particle of \( q \) Coulomb moving in a uniform electric field \( \vec{E} \). For the situation described above, obtain the equation of motion of the electron between the plates of the capacitor. (d) Hence show that the vertical downward displacement of the electron as a function of the time \( t \) it spends between the plates of the capacitor is \( \frac{q t^{2}}{m}|\vec{E}| \). (e) Use your answers above to find the electric field \( \vec{E} \) between the plates of the capacitor. \[ [4+3+3+3+5 \text { marks }] \]