Electricity and Magnetism
# Capacitors and Transformers

$+140.0\text{ pC}$ and $-140.0\text{ pC},$ respectively. If the potential difference between them is $10.0\text{ V},$ what is the capacitance of the system consisting of the two metal plates?

The two metal plates in the above figure have net charges of$40.0 \,\mu\text{F}$ and is initially uncharged. The battery provides a potential difference of $90.0\text{ V}.$ After switch **S** is closed, how much charge will pass through it?

Consider a spherical capacitor which consists of two concentric spherical shells of radii $38.0\text{ mm}$ and $40.0\text{ mm},$ respectively. If we want to make a parallel-plate capacitor with the same separation, what must be the area of each plate for the same capacitance?

The value of electrostatic constant is $\displaystyle k=\frac{1}{4\pi\varepsilon_0}=8.99 \times 10^9 \text{ N}\cdot\text{m}^2\text{/C}^2$ and the value of the permittivity constant is $\varepsilon_0=8.85 \times 10^{-12} \text{ C}^2\text{/N}\cdot\text{m}^2.$

One can make a homemade capacitor using an aluminum foil and a plastic or a wooden stick (it's important that the stick is an insulator). Make a small ball out of the aluminum foil and wrap it around the stick. Then make a wider foil sphere around the ball so that they don't touch, and your capacitor is ready!

If the radius of the smaller ball is $5~\mbox{cm}$ and of the bigger sphere $15~\mbox{cm}$, which capacitance **in pF** do you expect to obtain?

**Details and assumptions**

- The ball and the sphere are centered around the same point

If you want to make a parallel-plate capacitor of capacitance $0.9\text{ pF}$ with two metal plates with area $1.1\text{ cm}^2,$ what must be the separation between the two plates?

The value of the permittivity constant is $\varepsilon_0=8.85 \times 10^{-12} \text{ C}^2\text{/N}\cdot\text{m}^2.$