Relevant equations:

\(\displaystyle C=\frac{\varepsilon_{o}A}{D}\)

\(\displaystyle Q=CV\)

 \(\displaystyle U=\frac{1}{2}\frac{Q^{2}}{C}\)

Considering we are dealing with regions of charge instead of a point charge (or a charge that is at least spherical symmetric), it would be wise to consider using the definition of the electric field here:

\(\displaystyle E=\frac{V}{D}\)

The battery maintains a constant V so we can directly relate E and D. Since D is doubled, E will be halved. 

The voltage rise through the source must be the same as the drop through the capacitor.

\(\displaystyle V_1=V_{C1}\)

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

\(\displaystyle V_{C1}=E*d\)

Combine these equations and solve for the electric field:

\(\displaystyle V_1=E*d\)

\(\displaystyle \frac{V_1}{d}=E\)

Convert mm to m and plug in values:

\(\displaystyle \frac{24}{.0005}=E\)

\(\displaystyle E=4.8*10^4 \frac{N}{C}\)

The voltage rise through the source must be the same as the drop through the capacitor.

\(\displaystyle V_1=V_{C1}\)

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

\(\displaystyle V_{C1}=E*d\)

Combine equations and solve for the electric field:

\(\displaystyle V_1=E*d\)

\(\displaystyle \frac{V_1}{d}=E\)

Convert mm to m and plugging in values:

\(\displaystyle \frac{60V}{.001}=E\)

\(\displaystyle 60000\frac{N}{C}=E\)

Use the electric field in a capacitor equation:

\(\displaystyle E=\frac{q}{A*\epsilon_0}\)

Combine equations:

\(\displaystyle E*\epsilon_0*A=q\)

Converting \(\displaystyle cm^2\) to \(\displaystyle m^2\) and plug in values:

\(\displaystyle 60000*8.85*10^{-12}*.06=q\)

\(\displaystyle 31.9nC=q\)

The voltage rise through the source must be the same as the drop through the capacitor.

\(\displaystyle V_1=V_{C1}\)

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

\(\displaystyle V_{C1}=E*d\)

Combine equations and solve for the electric field:

\(\displaystyle V_1=E*d\)

\(\displaystyle \frac{V_1}{d}=E\)

Convert mm to m and plug in values:

\(\displaystyle \frac{V_1}{d}=E\)

\(\displaystyle 4000\frac{N}{C}=E\)

Since it is the only element in the circuit besides the source, the voltage drop across the capacitor must be equal to the voltage gain in the capacitor.

\(\displaystyle V_1=V_{C1}\)

Definition of voltage:

\(\displaystyle V=E*d\)

Combine equations:

\(\displaystyle V_1=E_{C1}*d\)

Solve for \(\displaystyle E_{C1}\)

\(\displaystyle \frac{V}{d}=E_{C1}\)

Convert to meters and plug in values:

\(\displaystyle \frac{5}{.0025}=E_{C1}\)

\(\displaystyle 2000\frac{N}{C}=E_{C1}\)

The voltage drop through the capacitor needs to be equal to the voltage of the battery.

\(\displaystyle V_b=V_c\)

The voltage drop of a parallel plate capacitor is equal to the internal electric field times the distance between them.

\(\displaystyle V_c=E*d\)

Combing equations and solving for \(\displaystyle E\)

\(\displaystyle E=\frac{V_b}{d}\)

From this, it can be seen that doubling the separation will halve the electric field.

The voltage drop through the capacitor needs to be equal to the voltage of the battery.

\(\displaystyle V_b=V_c\)

The voltage drop of a parallel plate capacitor is equal to the internal electric field times the distance between them.

\(\displaystyle V_c=E*d\)

Combing equations and solving for \(\displaystyle E\)

\(\displaystyle E=\frac{V_b}{d}\)

From this, it can be seen that doubling the voltage of the battery will doubled the electric field inside the capacitor.

The voltage rise through the source must be the same as the drop through the capacitor.

\(\displaystyle V_1=V_{C1}\)

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

\(\displaystyle V_{C1}=E*d\)

Combining equations:

\(\displaystyle V_1=E*d\)

Solving for \(\displaystyle E\):

\(\displaystyle \frac{V_1}{d}=E\)

Converting \(\displaystyle mm\) to \(\displaystyle m\) and plugging in values:

\(\displaystyle \frac{V_1}{d}=E\)

\(\displaystyle 6000\frac{N}{C}=E\)

Using the electric field in a capacitor equation:

\(\displaystyle E=\frac{q}{A*\epsilon_0}\)

Combining equations:

\(\displaystyle E*\epsilon_0*A=q\)

Converting \(\displaystyle cm^2\) to \(\displaystyle m^2\) and plugging in values:

\(\displaystyle (6000)*(8.86*10^{-12})*(4*10^{-4})=q\)

\(\displaystyle q=2.13*10^{-11}C\)

The voltage rise through the source must be the same as the drop through the capacitor. Since the batteries all have the same potential and are in parallel, the potential will be the same as if there was just one.

\(\displaystyle V_1=V_{C1}\)

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

\(\displaystyle V_{C1}=E*d\)

Combining equations:

\(\displaystyle V_1=E*d\)

Solving for E:

\(\displaystyle \frac{V_1}{d}=E\)

Converting \(\displaystyle mm\) to \(\displaystyle m\) and plugging in values:

\(\displaystyle \frac{V_1}{d}=E\)

\(\displaystyle 4000\frac{N}{C}=E\)

Using the electric field in a capacitor equation:

\(\displaystyle E=\frac{q}{A*\epsilon_0}\)

Combining equations:

\(\displaystyle E*\epsilon_0*A=q\)

Converting \(\displaystyle cm^2\) to \(\displaystyle m^2\) and plugging in values:

\(\displaystyle (4000)*(8.86*10^{-12})*(4*10^{-4})=q\)

\(\displaystyle q=1.42*10^{-11}C\)

Since all the resistors are connected in parallel, they will all experience the same voltage drop, which is equal to the voltage of the battery. Furthermore, since each resistor has the same resistance, the current flowing through the four resistors will be equal.

The current flowing through each resistor can be shown by Ohm's law:

\(\displaystyle V=iR\)

Also, the total current of the circuit can be found by noting:

\(\displaystyle V=i_{Total}R_{Total}\)

Moreover, the total resistance can be calculated by noting that resistors in parallel add inversely.

\(\displaystyle \frac{1}{R_{Total}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}\)

Since each resistor is the same, we can note that:

\(\displaystyle \frac{1}{R_{Total}}=\frac{4}{R}\)

\(\displaystyle R_{Total}=\frac{R}{4}\)

Plugging \(\displaystyle R_{Total}\) into the previous equation, we see that:

\(\displaystyle i_{Total}=\frac{V}{\frac{R}{4}}=\frac{4V}{R}\)

So in the case of four resistors connected in parallel, the total current in the circuit is equal to \(\displaystyle \frac{4V}{R}\) . And since the overall current is split evenly between the four resistors, we see that:

\(\displaystyle i=\frac{(\frac{4V}{R})}{4}=\frac{V}{R}\)

Now, if we consider the case of five resistors connected in parallel, we will see that the total resistance is:

\(\displaystyle \frac{1}{R_{Total}^{'}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}+\frac{1}{R_{5}}\)

\(\displaystyle \frac{1}{R_{Total}^{'}}=\frac{5}{R}\)

\(\displaystyle R_{Total}^{'}=\frac{R}{5}\)

Therefore, the overall current in the circuit when a fifth resistor is added in parallel is:

\(\displaystyle i_{Total}^{'}=\frac{V}{R_{Total}^{'}}=\frac{V}{(\frac{R}{5})}=\frac{5V}{R}\)

And once again, since each of the five resistors are connected in parallel, the total current will be split evenly between them, thus giving:

\(\displaystyle i^{'}=\frac{(\frac{5V}{R})}{5}=\frac{V}{R}\)

\(\displaystyle i=i^{'}=\frac{V}{R}\)

Therefore, we see that the current through each resistor is the same, both before and after the addition of the fifth resistor.

To summarize, the voltage drop across each resistor is the same since they are connected in parallel. Also, since each resistor has the same resistance, the current through each is the same. Adding a fifth resistor in parallel decreased the overall resistance of the circuit and increased the overall current flowing through the circuit. But because this new, greater current is divided evenly among all the resistors, including the new one that was added, the current through each of the resistors does not change.