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Electricity and Magnetism

Capacitors and Transformers



The two metal plates in the above figure have net charges of \(+140.0\text{ pC}\) 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 capacitor in the above figure has a capacitance 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 \(\varepsilon_0=8.85 \times 10^{-12} \text{ C}^2\text{/N}\cdot\text{m}^2.\)


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