Question

The diagram shows part of an evacuated tube in which a stream of electrons from an electron gun passes between a pair of parallel large deflecting plates. The vertical displacement of the electron beam as it leaves the parallel plates is $x$. (Do not consider gravity and the electron enters the deflecting region parallel to the plates)




Which of the following option(s) will increase the displacement $x$ of the beam as it leaves the parallel plates

• A. Increasing the accelerating voltage
• B. Increasing the deflecting voltage
• C. Increasing the distance between the electron gun and point of exit
• D. Increasing the distance between the two deflecting plates

Answer 1

Answer: (B), (C)

The correct options are
B Increasing the deflecting voltage
C Increasing the distance between the electron gun and point of exit




Due to accelerating voltage electron gains kinetic energy as shown. So,
$\frac{1}{2}m{u}^2=e{V}_a.........\left(1\right)(V_a:\text{accelerating voltage})$

Now, deflecting voltage causes deflecting force $F_x$ on electron.
$F_x=e\frac{V_d}{D}(V_d:\text{deflecting voltage)}$
$a_x=\frac{e{V}_d}{mD}$

From equation of kinematics
$x=\frac{1}{2}{a}_x{t}^2$
$x=\frac{e{V}_d{t}^2}{2mD}............\left(2\right)$
Here $t$ is time taken to achieve deflection of $x$.
In the same time particle travels a distance $L$ along $y$. So,
$y=L=ut$
Substituting $L$ in equation $\left(2\right)$
$\Rightarrow x=\frac{e{V}_d{L}^2}{2mD{u}^2}$
$\Rightarrow x=\frac{e{V}_d{L}^2}{4e{V}_{a}D}$ $(\because \frac{1}{2}m{u}^2=e{V}_a)$
$\Rightarrow x=\frac{V_d{L}^2}{4V_{a}D}$
$\Rightarrow x\propto \frac{1}{V_a}$, $x\propto V_d$ , $x\propto L^2$ , $x\propto \frac{1}{D}$