Capacitors and Capacitance (4/28/2015)

Quiz

We were hit with a surprise quiz today! The question is pictured above, and we had to solve for every unknown in the problem, which were the currents and power.

Shown above are Professor Mason's and our table's calculations for the problem. We solved for the currents I1, I2, and I3, and power P1 and P2 being used in the circuit. In the photo on the right, the value of I2 is a negative number. We cannot have negative current, so the negative sign is actually direction, telling us that the way we placed I2 is incorrect and it should be flowing in the opposite direction.

Energy Storage

Above is the derived and modified formula of the energy stored in a capacitor. The work it takes to move a differential charge dq through a voltage V is V dq. Integrating from when there is no charge on the capacitor until the capacitor is fully charged with a charge Q we get:

deltaU = W = (integral from 0 to Q) V dq. 

If you continue to integrate that out, and replace Q with CV, you end up with the equation above.

Predicting Dependence on Area and Separation

Depicted above is the fact that capacitance (a measure of the ability of a device to store charge per unit of voltage applied across the device) is proportional to the area of the conductors in a capacitor, and inversely proportional to the distance between the plates. A capacitor itself is any two conductors separated by an insulator which can be electrically charged so that one conductor has a positive charge and the other conductor has an equal amount of negative charge.

An example of a charging capacitor, the formula for capacitance, and the units of capacitance (F).

Measuring How Capacitance Depends on Area or Separation

In order to verify the prediction above, we were asked to create our own capacitors from which we would gather data and plot a graph so that we can see the relationship between capacitance and area/distance.

What we ended up doing was separating two foil sheets with varying sheets of paper. We placed the foil in a notebook, separated it by paper pages, and closed the book in order to keep the book as closed as possible. Above you can see the results of capacitance, which was inversely proportional to the number of pages used (the distance apart). We did not conduct a test to see if capacitance was proportional to area, but logically we can expect that since a larger surface can hold more charges.

Our graph depicting the inverse relationship between capacitance and distance between conductors.

Deriving a Mathematical Expression for Capacitance

Given that Q = eEA = ke₀EA and that in a uniform electric field the voltage drop over a distance d is given by V = Ed, we can use what we know about charges in conductors and the relationship between potential difference, V, and electric field to derive an expression for the capacitance, C, of a parallel plate capacitor in terms of the area, A, and separation, d, of the aluminum plates. Above, you can see the results of plugging in the known values into the equation C = Q/V.

To test this newly derived equation, we were asked that if we have two square foil sheets, separated by paper with a dielectric constant of 3.5 that is 1 mm thick, how long (in miles) would each side of the sheets have to be in order to have C = 1 F? The answer shown above is incorrect. The answer for the area is 32284100 meters-squared, which square-rooted becomes 5681.9 m. After converting to miles you get 3.53 miles. In order to make a circuit element smaller, while still having this same capacitance would be to use a material with a better (larger) dielectric constant.

Capacitors in Series and Parallel

For this test, we took the capacitance of two capacitors individually, and in two different configurations. What we see from the data above, is that the capacitance of capacitors in series add inversely as 1/C1 + 1/C2 = 1/F. Alternatively, capacitors in parallel add directly as C1 + C2 = F.

Equivalent or Effective Capacitance

To test this capacitor addition, we solved the problem above by first adding the capacitors in series, and then adding all the resulting capacitance values in parallel, ending up with 6.5 uF.

Capacitors in a Circuit Problem 

For the final activity, we were given the oddly shaped circuit shown above, where three capacitors are connected to a battery of unknown voltage. Professor Mason's interpretation of the circuit is shown above, while ours is shown below where we proceeded to calculate the unknown voltage Vb.

We started by solving for the Q of each capacitor since Q = CV, and we were given the voltage drop at V₁ = 11 V. Since the top capacitors, C₁ & C₂, are in series we were able to calculate V₂, since they must both have the same charge. We now have the Q of the top capacitors, and know that their added capacitance is 12nF. We can now determine the voltage drop across the effective capacitance C₁₂ which is also equal to the voltage of the battery using the equation V_b = V₁₂ = Q₁₂/C₁₂= 220nC/12nF = 18.3 V.

We then go on to calculate the work done by the battery to charge up the system of capacitors.

The work needed to charge up the capacitors is equal to the energy stored in each of the capacitors, so we use the equation Ut = U1 + U2 + U3, where U = 1/2*C*V^2. As seen above, we did this in one fell swoop by using the total C and total V of the circuit, and ended up with a rounded answer of 9x10^-6 J.

Summary: 

In summation, we reviewed the basics of how to calculate current and power in a circuit using Kirchhoff's Laws. We also learned about capacitance and how it functions in a circuit, by seeing how changing different variables affects capacitance, and how to compute total capacitance of circuits. Our final task was a culmination of what we learned about capacitance, as well as recalling various rules about potential difference through a circuit.