The two capacitors depicted in the figure consist of square plates of edge a . Their plate separations are d₁ and d₂ respectively. A potential difference V is applied across the points a and b . An electron is projected along the central line between the plates of the upper capacitor. At what minimum speed must the electron be projected to ensure it does not collide with either plate? Account for electric forces only

Step-by-step solution

The two capacitors are connected in series. Let their capacitances be C₁ = ₀ a^2 d₁ and C₂ = ₀ a^2 d₂ . The potential difference across the upper capacitor is given by V₁ = V C₂ C₁ + C₂ . Substituting the expressions for capacitance, we get: V₁ = V ₀ a^2 d₂ ₀ a^2 d₁ + ₀ a^2 d₂ = V d₁ d₁ + d₂ The electric field between the plates of the upper capacitor is: E₁ = V₁ d₁ = V d₁ + d₂ The acceleration of the electron in the vertical direction is: a_y = e E₁ m = e V m (d₁ + d₂) The time taken by the electron to cross the p