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.
Figure

Answer 1

Explanation:

Step 1:

Consider:

Electron’s velocity = u

Electron’s mass = m

So, the horizontal distance is measured as $x=u \times t$

So, $t=x v$ …(i)

Step 2:

Inside the capacitor, the electric field is E.

Therefore, electron’s acceleration is qEm.

Vertical distance,  

$y=12 \text { q } E m t 2$ Therefore, $t=x u$

$y=d 12 \text { and } x=a \Rightarrow d 12=12 \text { qEm } \cdot \text { au } 2$    …(ii)

Step 3:

Two capacitors’ capacitance:

$C_{1}=€ 0 \mathrm{a} 2 \mathrm{d} 1 \text { and } C_{2}=\in 0 \mathrm{a} 2 \mathrm{d} 2$

It is specified that the connection of capacitors is in series. So, the equivalent capacitance is shown as  

Ceq = C2 Ceq + C1C2C1 = ∈0 a2d2∈0 a2d1 x C1C2C1 + ∈0 a2d2 = ∈0 a2d1 + d2

Step 4:

On the capacitors systems, the total charge is Q = CeqV = ∈0 a2d1+d2V

There are series of capacitors, so the charge is same on the both.

Across the capacitor, the potential difference is given as V = Vd1d1 + d2.

Inside the capacitor, the electric field’s magnitude is shown as E = Vd1 + d2.

The charge on electron q is represented by e.

When the values of E and q in (ii), we obtain

$\Rightarrow \mathrm{d} 12=12 \mathrm{gVm}(\mathrm{d} 1+\mathrm{d} 2) \cdot \mathrm{au} 2$ …(iii)

The electron’s minimum velocity is shown as  

$U=V e a 2 m d 1 d 1+d 21 / 2$