Equivalent capacitance in parallel is calculated by taking the sum of each individual capacitor. We can reduce the two parallel capacitors as the following:

$\displaystyle C_{eq}=C2+C3 = 7F+1F= 8F$

The new equivalent circuit has two capacitors in series. This requires us to sum the reciprocals to find equivalent capacitance:

$\displaystyle \frac{1}{C_{tot}}=\sum\frac{1}{C} = \frac{1}{5}+\frac{1}{8} = \frac{13}{40}$

$\displaystyle C_{tot} = 3.08F$

To find the total capacitance of capacitors in series, you use the following equation:

$\displaystyle \frac{1}{C_{tot}} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+...+\frac{1}{C_n}$.

Our values for $\displaystyle C_1$, $\displaystyle C_2$, and $\displaystyle C_3$ are 4, 3, and 2. Now, we can plug in our values to find the answer.

$\displaystyle \frac{1}{C_{tot}} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$\displaystyle \frac{1}{C_{tot}}=\frac{1}{4}+\frac{1}{3}+\frac{1}{2}$

$\displaystyle \frac{1}{C_{tot}}= \frac{13}{12}$

The answer we have is the inverse of this. Therefore, the total capacitance is $\displaystyle \frac{12}{13} \mu F$.

Remember, the equations for adding capacitances are as follows:

$\displaystyle (Series)\ \frac{1}{C_{tot}}&=\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_n}$

$\displaystyle (Parallel)\ C_{tot}&=C_1+C_2+...+C_n$

Capacitors $\displaystyle A$ and $\displaystyle B$ are in series, $\displaystyle AB$ and $\displaystyle C$ are in parallel, and $\displaystyle ABC$ and $\displaystyle D$ are in parallel.

$\displaystyle C_{AB}&=(\frac{1}{C_A}+\frac{1}{C_B})^{-1}$

$\displaystyle C_{AB}=(\frac{1}{3}+\frac{1}{6})^{-1}$

$\displaystyle C_{AB}= 2$

$\displaystyle C_{ABC}=C_{AB}+C_C$

$\displaystyle C_{ABC}=2+4$

$\displaystyle C_{ABC}= 6$

$\displaystyle C_{total}=C_{ABC}+C_{D}$

$\displaystyle C_{total}=6+1$

$\displaystyle C_{total}=7$

The key hint to remember here is that the capacitance depends only on the geometry of the material, not the potential difference or electric field.

For a parallel-plate capacitor,

$\displaystyle C=\frac{\epsilon _{0}A}{d}$

Plugging in the numbers given results in 

$\displaystyle C=\frac{8.854\cdot 10^{-12}\frac{C^2}{N\cdot m^2}\cdot0.0015m^2}{0.03m}$

$\displaystyle C=4.43\cdot 10^{-13}F$

Be careful to convert the units to meters!

Capacitors in parallel combine according to the following equation:

$\displaystyle C_{tot}=C_1+C_2+...+C_n$.

Because the capacitances are additive, and all of the capacitances are greater than zero, no matter what numbers you use, you will always end up with a number that is greater than any of the individual numbers.

Let's go through all of them and find the equivalent capacitance.

(A) This is just 

$\displaystyle C_{eq}=C$

(B) There are two capacitors in series, so this is 

$\displaystyle C_{eq}=\left ( \frac{1}{C}+\frac{1}{C}\right )^{-1}=\left ( \frac{2}{C}\right )^{-1}=\frac{1}{2}C$

(C) These capacitors are in parallel, so

$\displaystyle C_{eq}=C+C+C=3C$

(D) These are a combination of series and parallel.  Two are in series and they are in parallel with a third,

$\displaystyle C_{eq}=\left ( \frac{1}{C}+\frac{1}{C}\right )^{-1}+C=\left ( \frac{2}{C}\right )^{-1}+C=\frac{3}{2}C$

So, ranking them we get

$\displaystyle C>D>A>B$

In this question, we're told the maximum amount of charge that a capacitor can hold at a given voltage. We're then asked to determine the capacitance. To do this, we'll need to use the expression for capacitance.

$\displaystyle C=\frac{Q}{V}$

Plug in the values given to us in the question stem:

$\displaystyle C=\frac{36 C}{12 V}=3F$

The fact that the system is still connected to the battery indicate a constant V so regardless what happens to the capacitor, V stays fixed.

Relevant equations:

$\displaystyle Q=CV$

$\displaystyle C=\frac{\varepsilon_{o}A}{D}$

Plug the second equation into the first:

$\displaystyle Q=\frac{\varepsilon _{o}AV}{D}$

Considering all the variables in the numerator are held fixed for this problem, we see that increasing D will decrease the charge stored on each plate. 

The charge has no where to go. Without the battery connected, the charge has no physical avenue on or off the plates.