Solution
1. Charge Leakage:
The ball is placed in a conducting medium with resistivity $\rho$ and permittivity $\varepsilon = \varepsilon_r \varepsilon_0$. A conducting medium cannot sustain a static free charge distribution; the charge leaks away into the medium. By applying Gauss’s Law and the Continuity Equation to the space surrounding the ball:
The leakage current $I$ flowing out from the ball is related to the instantaneous charge $q(t)$ by:
$$ I = \oint \vec{J} \cdot d\vec{A} = \oint \frac{\vec{E}}{\rho} \cdot d\vec{A} = \frac{1}{\rho} \oint \vec{E} \cdot d\vec{A} $$Using Gauss’s Law $\oint \vec{E} \cdot d\vec{A} = q(t)/\varepsilon$:
$$ I = \frac{q(t)}{\rho \varepsilon} $$Since current is the rate of loss of charge ($I = -dq/dt$):
$$ \frac{dq}{dt} = -\frac{q}{\rho \varepsilon} $$Integrating this gives the charge as a function of time:
$$ q(t) = q_0 e^{-t/\tau} \quad \text{where } \tau = \rho \varepsilon = \rho \varepsilon_r \varepsilon_0 $$2. Equation of Motion:
The forces acting on the ball are the electric force $F_E$ driving it forward and the viscous drag force $F_v$ opposing the motion. The electric field inside the capacitor is uniform: $E_{ext} = V/d$.
$$ F_{net} = F_E – F_v = q(t)E_{ext} – \beta v $$ $$ ma = q(t)\frac{V}{d} – \beta v $$Since the ball is described as “almost massless”, we can take the limit $m \to 0$. This implies the terminal velocity is reached instantaneously at every moment:
$$ 0 \approx q(t)\frac{V}{d} – \beta v $$ $$ v(t) = \frac{V}{\beta d} q(t) $$3. Calculating Maximum Distance:
Substitute $q(t)$ into the velocity equation:
$$ v(t) = \frac{V}{\beta d} q_0 e^{-t/\tau} $$The velocity $v = dx/dt$. To find the total distance $x_{max}$ traveled before the ball stops (as $t \to \infty$), we integrate:
$$ x_{max} = \int_{0}^{\infty} v(t) dt = \frac{V q_0}{\beta d} \int_{0}^{\infty} e^{-t/\tau} dt $$Evaluating the integral $\int_{0}^{\infty} e^{-t/\tau} dt = \tau$:
$$ x_{max} = \frac{V q_0}{\beta d} \cdot \tau $$Substituting $\tau = \rho \varepsilon_r \varepsilon_0$: