Question

A parallel plate capacitor with plates of length $l$ is included in a circuit as shown in figure. Given are the emf of the current source, its internal resistance $r$ and the distance $d$ between the plates. An electron with a velocity $V_0$ flies into the capacitor, parallel to the plates as shown. Find the value of resistance $R$ (in $\Omega$) which should be connected in parallel with the capacitor so that the electron flies out of the capacitor at an angle of $37^0$ to the plates. (Assume that circuit is in steady state).

Answer 1

The question is incomplete as it does not provide numerical values for the given parameters ($\mathcal{E}$, $r$, $l$, $d$, $V_0$). Therefore, a numerical value for $R$ cannot be calculated. The derived formula for $R$ is: $R = \frac{3 m_e d V_0^2 r}{4 e l \mathcal{E} - 3 m_e d V_0^2}$

Answer 2

1. Steady State Analysis: In steady state DC, a capacitor acts as an open circuit. Thus, the current from the source ($\mathcal{E}$ with internal resistance $r$) flows entirely through the parallel resistance $R$. The total resistance in the circuit is $R+r$. The current is $I = \frac{\mathcal{E}}{R+r}$. The voltage across the capacitor ($V_{cap}$) is equal to the voltage across $R$, which is $V_{cap} = I \cdot R = \frac{\mathcal{E}R}{R+r}$.

2. Electron Motion Analysis: The electron enters the capacitor with horizontal velocity $V_0$. The electric field $E$ between the plates is $E = \frac{V_{cap}}{d}$. The force on the electron is $F_y = eE = \frac{eV_{cap}}{d}$ (upwards, as the electron is negative and the upper plate is positive). The acceleration of the electron in the vertical direction is $a_y = \frac{F_y}{m_e} = \frac{eV_{cap}}{m_e d}$. The time of flight $t$ through the capacitor (length $l$) is $t = \frac{l}{V_0}$. The vertical velocity at the exit is $v_y = a_y t = \frac{eV_{cap}}{m_e d} \cdot \frac{l}{V_0}$.

3. Exit Angle Condition: The electron exits at an angle of $37^0$ to the plates. This means the ratio of vertical velocity to horizontal velocity at exit is $\tan 37^0$. $\frac{v_y}{V_0} = \tan 37^0 = \frac{3}{4}$. So, $v_y = \frac{3}{4} V_0$.

4. Solving for R: Equating the two expressions for $v_y$: $\frac{eV_{cap}}{m_e d} \cdot \frac{l}{V_0} = \frac{3}{4} V_0$ $V_{cap} = \frac{3 m_e d V_0^2}{4 e l}$

   Now, equate this with the expression for $V_{cap}$ from the circuit analysis: $\frac{\mathcal{E}R}{R+r} = \frac{3 m_e d V_0^2}{4 e l}$

   Let $K = \frac{3 m_e d V_0^2}{4 e l}$. Then, $\frac{\mathcal{E}R}{R+r} = K$. $\mathcal{E}R = K(R+r) = KR + Kr$ $R(\mathcal{E} - K) = Kr$ $R = \frac{Kr}{\mathcal{E} - K}$ Substituting the value of $K$: $R = \frac{\left(\frac{3 m_e d V_0^2}{4 e l}\right) r}{\mathcal{E} - \left(\frac{3 m_e d V_0^2}{4 e l}\right)} = \frac{3 m_e d V_0^2 r}{4 e l \mathcal{E} - 3 m_e d V_0^2}$