The relationship between capacitance, distance, and area is \(\displaystyle C=\epsilon_0 \frac{A}{d}\). We can rearrange this equation to solve for area.

\(\displaystyle A=\frac{Cd}{\epsilon_0}\)

Now, we can use the values given in teh question to solve.

\(\displaystyle A=\frac{1F(1*10^{-2}m)}{8.85* 10^{-12} \frac{F}{m}}=1.1*10^9m^2\)

The electric potential difference created between the plates of a parallel plate capactor is given by the equation:

 \(\displaystyle V = \frac{Q}{C}\) 

The charge can be calculated by using the equation:

\(\displaystyle Q=\sigma A\)

The value of the capacitance is related to the dimensions of the capacitor with the equation:

\(\displaystyle C=\kappa \frac{\varepsilon _{o}A}{d}\)

Combining these equations yields:

\(\displaystyle V=\frac{\sigma A}{\frac{\kappa \epsilon_0A}{d}}=\frac{\sigma d}{\varepsilon_{o} \kappa }\)

The area becomes inconsequential, while the potential is directly proportional to the surface charge density and the distance between the plates, and inversely proportional to the dielectric of the material between the plates. Changing the area does not cause any change in the potential difference measured between the plates, and changing any of the other variables would cause a resultant change in the potential difference.

For parallel plate capacitors, the equation is \(\displaystyle C=\frac{\epsilon_0A}{d}\).

Solve for \(\displaystyle A\).

\(\displaystyle A=\frac{dC}{\epsilon_0}\)

Now we can plug in our given values.

\(\displaystyle A=\frac{(0.5*10^{-3})(700*10^{-6})}{8.85*10^{-12}\frac{\text{C}^2}{\text{Nm}^2}}\)

\(\displaystyle A=3.95*10^4\ \text{m}^2\)

Capacitance is related to plate area and distance by the equation \(\displaystyle C=\epsilon_0\frac{A}{d}\).

Given the area and distance, we can solve for capacitance. The voltage, in this case, is irrelevant.

\(\displaystyle C=8.85* 10^{-12} \frac{F}{m}(\frac{5 m^2}{3*10^{-3}m})=8.85* 10^{-12} \frac{F}{m}(1.67*10^{3}m)=1.5*10^{-8}F\)

Charge on a capacitor is given by the equation \(\displaystyle Q=CV\). We know that the voltage is \(\displaystyle 1000V\), but we need to determine the capacitance based off of the area and distance between the plates.

\(\displaystyle C=\epsilon_0\frac{A}{d}\)

\(\displaystyle C=(8.85*10^{-12}\frac{F}{m})(\frac{5m^2}{3*10^{-3}m})\)

\(\displaystyle C=(8.85*10^{-12}\frac{F}{m})(1.67*10^{3}m)=1.5*10^{-8}F\)

We can plug this value into the equation for charge.

\(\displaystyle Q=CV=(1.5*10^{-8}F)(1000V)\)

\(\displaystyle Q=1.5*10^{-5}C\)

The electric field given by a capacitor is given by the formula \(\displaystyle E=\frac{Q}{\epsilon_0 A}\). We do not have these variables, so we will have to adjust the equation.

\(\displaystyle Q=CV\rightarrow E=\frac{CV}{\epsilon_0A}\)

The capacitance can be determined by the area of the plates and the distance between them.

\(\displaystyle C=\epsilon_0\frac{A}{d}\rightarrow E=\frac{\epsilon_0\frac{A}{d}V}{\epsilon_0A}\)

This equation simplifies to \(\displaystyle E=\frac{V}{d}\), allowing us to solve using the values given in the question.

\(\displaystyle E=\frac{1000V}{3*10^{-3}m}=3.33*10^{5}\frac{N}{C}\)

Relevant equations:

\(\displaystyle \frac{1}{C_{eq, series}}=\frac{1}{C_1}+\frac{1}{C_2}+ ...\)

Use the series equation, replacing C1 and C2 with the given constant C:

\(\displaystyle \frac{1}{C_{eq, series}}= \frac{1}{C}+ \frac{1}{C}\)

\(\displaystyle \frac{1}{C_{eq, series}}=\frac{2}{C}\)

\(\displaystyle C_{eq, series} = \frac{C}{2}\)

This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.

Relevant equations:

\(\displaystyle W = q\Delta V\)

\(\displaystyle W = \Delta KE\)

\(\displaystyle C = \frac{Q}{V}\)

\(\displaystyle C = \frac{\epsilon _o A}{d}\)

According to the work-energy theorem, work done on the proton is equal to its change in kinetic energy. 

1. Find an expression for potential difference, \(\displaystyle V\), in terms of \(\displaystyle A\) and \(\displaystyle d\), by setting the two capacitance equations equal to each other:

\(\displaystyle \frac{Q}{V}=\frac{\epsilon _o A}{d}\)

\(\displaystyle V = \frac{Qd}{\epsilon _o A}\)

2. Multiply the potential difference times the proton charge to find work (and thus kinetic energy):

\(\displaystyle W = \frac {eQd}{\epsilon _o A} = \Delta K\)

Relevant equations:

\(\displaystyle Q = CV\)

\(\displaystyle \frac{1}{C_{eq, series}} = \frac{1}{C_1} + \frac{1}{C_2}\)

First, find the equivalent capacitance of the whole circuit:

\(\displaystyle \frac{1}{C_{eq,series}}= \frac{1}{5} + \frac{1}{10} = \frac{3}{10}\)

\(\displaystyle C_{eq, series} = \frac{10}{3} \mu F\)

Use this equivalent capacitance to find the total charge:

\(\displaystyle Q = CV = (\frac{10}{3} \mu F) * (12 V) = 40 \mu C\)

Capacitors in series always have the same charge on each unit, so the charge on the \(\displaystyle 5 \mu F\) capacitor is also \(\displaystyle 40 \mu C\).