In the setup shown, when a particle P of charge q and mass m enters with velocity u in an uncharged parallel plate capacitor of capacitance C, the switch is closed. If time spent by the charge particle inside the capacitor is negligible $(t < < < < R C)$ as compared to the time constant RC of the circuit, then acceleration of the particle as function of time t is ____ (Assume that the particle neither collides with the plates nor affects charge distribution on the plates, also, neglect the edge effects and gravity)
$[\begin{array}{l} t a k e e^{x} = 1 + x \\ w h e n x i s v e r y s m a l l \end{array}]$

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$\frac{q v t}{m d R C}$

$\frac{2 q v t}{m d R C}$

$\frac{q v t}{4 m d R C}$

$\frac{q v t}{8 m d R C}$

answer is A.

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Detailed Solution

$V_{C} = \frac{q}{C} = \frac{q_{0}}{C} (1 - e^{\frac{- t}{τ}}) = V (1 - e^{\frac{- t}{τ}}) E = \frac{V}{d} a = \frac{q E}{m} = \frac{q V_{C}}{m d} = \frac{q V}{m d} (1 - e^{\frac{- t}{τ}}) t < < τ a = \frac{q V t}{m d τ} = \frac{q V t}{m d R C}$