Water and Electricity

We know that $I=\frac{\mbox{d} Q}{\mbox{d} t}=\frac{\mbox{d} (CV)}{\mbox{d} t}=V\hspace{1mm} \frac{\mbox{d} C}{\mbox{d} t}+C\hspace{1mm} \frac{\mbox{d} V}{\mbox{d} t}$ Since $V=120\mbox{V}$ is constant, the current can simply be expressed as $I=120\times \frac{\mbox{d} C}{\mbox{d} t}$ Thus, we need to find an expression for the total capacitance $C$ as a function of time.

It is important to realise that since one part of the capacitor is immersed in water while the other part is exposed to air, the two parts can be treated as two different capacitors. Furthermore, by examining how the voltage source is connected to these two capacitors closely, we would realise that the two capacitors are connected in parallel series.

Next, let the capacitance of the capacitor exposed to air and the capacitor exposed to water be $C_1$ and $C_2$ respectively. Then, $C=C_1+C_2$ Now, let the distance between the capacitor plates, width, total height, and height of part of the capacitor immersed in water be $D$ , $x$ , $L$ , and $y$ respectively. $C_1$ and $C_2$ can then be expressed as $C_1=\epsilon_0 \frac{x (L-y)}{D}$

$C_2=\epsilon \hspace{1mm} \epsilon_0 \frac{xy}{D}$

However, since $C_0$ can be expressed as $C_0=\epsilon_0 \frac{xL}{D}$

We can rewrite the expressions for $C_1$ and $C_2$ as $C_1=\left( 1-\frac{y}{L} \right) C_0$

$C_2=\left( \frac{y}{L} \epsilon \right) C_0$

Meanwhile, we also realise that $\frac{y}{L}$ can be expressed in terms of time, as follows: $\frac{y}{L} =\frac{vt}{v\tau } =\frac{t}{20}$

Therefore, we can rewrite the expressions for $C_1$ and $C_2$ again in terms of time. Adding the two expressions would give the following expression of $C$ in terms of $t$ : $C=\left[ 1+\frac{t}{20} (\epsilon -1) \right]C_0$

In conclusion, the value of the current can be obtained as follows: $I=120\times \frac{\mbox{d} C}{\mbox{d} t} =120\times \left( \frac{\epsilon -1}{20} \right) C_0=5.76\times 10^{-3} \mbox{A}$

The wording was kind misleading in a way. The question should have specified that the average current was asked. I infered it from the fact that it was asking for the current for a finite time interval.

The charge stored initially on the capacitor is $C_0 V$ , and charge stored on it finally is $\epsilon C_0 V$ . So net charge, flown through the battery is $Q= (\epsilon -1)C_0 V$ . Thus average current is $\frac Q\tau= \frac{ (\epsilon -1)C_0 V}\tau$

$\quad \quad \quad \quad I\quad =\quad \frac { \Delta Q }{ t } \\ \\ \Rightarrow \quad I\quad =\quad \frac { { C }_{ f }V\quad -\quad { C }_{ i }V }{ t } \\ \\ \Rightarrow \quad I\quad =\quad \frac { 81{ C }_{ i }V\quad -\quad { C }_{ i }V }{ t } \\ \\ \Rightarrow \quad I\quad =\quad \frac { 80{ C }_{ i }V }{ t } \\ \\ \Rightarrow \quad I\quad =\quad \frac { 80\quad \times \quad 12\quad \times \quad { 10 }^{ -6 }\quad \times \quad 120 }{ 20 } \\ \\ \Rightarrow \quad I\quad =\quad 0.006$