Question

Both the capacitors shown in figure (31-E12) are made of square plates of edge a. The separations between the plates of the capacitors are d₁ and d₂ as shown in the figure. A potential difference V is applied between the points a and b. An electron is projected between the plates of the upper capacitor along the central line. With what minimum speed should the electron be projected so that it does not collide with any plate? Consider only the electric forces.

Answer 1

Let:
Velocity of the electron = u
Mass of the electron = m
Now,
Horizontal distance, x = u × t
$\Rightarrow t=\frac{x}{v}$ ...(i)

Let the electric field inside the capacitor be E
∴ Acceleration of the electron =$\frac{qE}{m}$
Vertical distance, $y=\frac{1}{2}\frac{qE}{m}{t}^2=\frac{1}{2}\frac{qE}{m}{\left(\frac{x}{u}\right)}^2\left(\because t=\frac{x}{u}\right)$
$$\begin{gathered} y=\frac{d_1}{2}\mathrm{and}x=a \\ \Rightarrow \frac{d_1}{2}=\frac{1}{2}\frac{qE}{m}·{\left(\frac{a}{u}\right)}^2...\left(\mathrm{ii}\right) \end{gathered}$$

Capacitance of the two capacitors:
C₁ = $\frac{{\in }_0{a}^2}{d_1}$ and C₂ = $\frac{{\in }_0{a}^2}{d_2}$
It is given that the capacitors are connected in series.
Thus, the equivalent capacitance is given by
$C_{\mathrm{eq}}=\frac{C_1{C}_2}{C_1+C_2}$

$C_{\mathrm{eq}}=\frac{{\displaystyle \frac{{\in }_0{a}^2}{d_1}}\times {\displaystyle \frac{{\in }_0{a}^2}{d_2}}}{{\displaystyle \frac{{\in }_0{a}^2}{d_1}}+{\displaystyle \frac{{\in }_0{a}^2}{d_2}}}=\frac{{\in }_0{a}^2}{\left(d_1+d_2\right)}$

Total charge on the system of capacitors, Q = C_eqV = $\frac{{\in }_0{a}^2}{\left(d_1+d_2\right)}V$
As the capacitors are in series, charge on both of them is the same.

The potential difference across the capacitor containing the electron is given by
$V=\frac{Q}{C_1}=\frac{{\in }_0{a}^{2}V}{C_1\left(d_1+d_2\right)}=\frac{{\in }_0{a}^{2}V}{\left({\displaystyle \frac{{\in }_0{a}^2}{d_1}}\right)\left(d_1+d_2\right)}=\frac{V{d}_1}{d_1+d_2}$
The magnitude of the electric field inside the capacitor is given by
$E=\frac{V}{d_1}=\frac{V}{d_1+d_2}$
The charge on electron q is represented by e.
On putting the values of q and E in (ii), we get
$\Rightarrow \frac{d_1}{2}=\frac{1}{2}\frac{qV}{m(d_1+d_2)}·{\left(\frac{a}{u}\right)}^2...\left(\mathrm{iii}\right)$

The minimum velocity of the electron is given by
$u={\left(\frac{Ve{a}^2}{m{d}_1\left(d_1+d_2\right)}\right)}^{1/2}$