Both the cacpacitors shown in figure are made ofsquare of edge a The separations between the plates of the capacitors are d_1and d_2as shown in the figure A potential difference V is applied between the point a and b .An electron is projected between the plate s of the upper capacitor along the central line . With what minimum speed should the electron be projected so that it is does not collide with any plate ? consider only the electric forces.

Explanation

Let the velocity of elctron = V ,
mass of electron = m
Equivalent capacitance = C`
`((eplison_0a^(2)/(d_1)xx(epsilon_0 a^2))/(d_2)/((eplison_0a^(2)/d_1)+(epsilon_0 a^2))
`= eplision_0 a^(2) / (d_1 + d_2)`
(because capacitors are in series)
Horizontal distance = V xx t`
Verticle distance , `y = 1 / 2 at^(2)`
Accelertion of electron `a = qE / m`
`y = d_1`
So, `d_1 = 1 /2 qE / m t^(2) `
or ` d_1 = 1 / 2 qE / m (x /v ) ^(2) `
or, `q = CV`
Put the value of the equation (i) , q , E , we get ,
minimum velocity at electron
`= ( V qa^(2) / md_1 (d_1 - d_2) ^(1/2)`