Solution
Analysis of Capacitor Charging:
When the switch is closed at $t=0$, the capacitor $C$ begins to charge through the resistor $R$. The charge on the capacitor plates $q_c(t)$ as a function of time is given by:
$$ q_c(t) = CV(1 – e^{-t/RC}) $$We are given that the time spent by the particle inside the capacitor is negligible compared to the time constant $RC$. This implies we are observing the system at very early times where $t \ll RC$. Using the small angle approximation $e^{-x} \approx 1 – x$ for small $x$:
$$ q_c(t) \approx CV \left[ 1 – \left( 1 – \frac{t}{RC} \right) \right] = CV \frac{t}{RC} = \frac{Vt}{R} $$Alternatively, this result can be understood by noting that at $t=0$, the uncharged capacitor acts as a short circuit, so the initial current is $I \approx V/R$. Since the time is short, the current is approximately constant, and charge $q_c \approx I \cdot t = \frac{V}{R}t$.
Electric Field and Acceleration:
The potential difference across the capacitor plates at time $t$ is:
$$ V_c(t) = \frac{q_c(t)}{C} = \frac{Vt}{RC} $$The electric field $E$ between the plates (separated by distance $d$) is:
$$ E(t) = \frac{V_c(t)}{d} = \frac{Vt}{d RC} $$The particle P carrying charge $q$ and mass $m$ experiences a force $F = qE$ inside the capacitor. By Newton’s second law, the acceleration $a$ is:
$$ a(t) = \frac{F}{m} = \frac{qE(t)}{m} $$Substituting the expression for $E(t)$:
$$ a(t) = \frac{q}{m} \left( \frac{Vt}{d RC} \right) $$