Question 10 Solution

The beam acts as a constant current source charging the sphere. The current $I_{in}$ is given by the rate at which charge hits the cross-sectional area of the sphere ($A = \pi r^2$). $$ I_{in} = n e (\pi r^2) v $$

Let $q$ be the charge on the sphere and $i$ be the current flowing to ground.
Potential of sphere: $V = q/C$, where $C = 4\pi\epsilon_0 r$.
Voltage across inductor: $V_L = L \frac{di}{dt}$.
Equating potentials: $\frac{q}{C} = L \frac{di}{dt}$.
Charge balance: $\frac{dq}{dt} = I_{in} – i$.

Differentiating the potential equation: $$ \frac{1}{C} \frac{dq}{dt} = L \frac{d^2i}{dt^2} $$ Substitute $\frac{dq}{dt}$: $$ \frac{I_{in} – i}{C} = L \frac{d^2i}{dt^2} \implies \frac{d^2i}{dt^2} + \frac{1}{LC}i = \frac{I_{in}}{LC} $$ Solution for $i(0)=0$: $$ i(t) = I_{in}(1 – \cos(\omega t)), \quad \omega = \frac{1}{\sqrt{LC}} $$

Max Current ($I_{max}$):
$i(t)$ oscillates between $0$ and $2I_{in}$. $$ I_{max} = 2 I_{in} = 2 n e \pi r^2 v $$

Max Charge ($Q_{max}$):
$q = LC \frac{di}{dt} = LC (I_{in}\omega \sin(\omega t))$.
$Q_{max} = LC I_{in} \omega = I_{in} \sqrt{LC}$.
Substituting values: $$ Q_{max} = (n e \pi r^2 v) \sqrt{L (4\pi\epsilon_0 r)} = 2 n e \pi r^2 v \sqrt{\pi \epsilon_0 r L} $$