Question 16.4: A lecture-demonstration capacitor consists of two parallel c...

$\text { (a) The circular plates have area } A=\pi r^{2}$. Then the capacitance is

$C=\frac{\varepsilon_{0} A}{d}=\frac{\left(8.85 \times 10^{-12} F / m \right) \cdot \pi(0.12 m )^{2}}{1.0 \times 10^{-4} m }$.

$=4.0 \times 10^{-9} F =4.0 nF$.

(b) Solving for d gives

$d=\frac{\varepsilon_{0} A}{C}=\frac{\left(8.85 \times 10^{-12} F / m \right) \cdot \pi(0.12 m )^{2}}{1.0 \times 10^{-6} F }$.

$=4.0 \times 10^{-7} m$.

That’s extremely close, and even a modest potential difference will result in sparks jumping and discharging the capacitor.

REFLECT Small values of capacitance are typical, often measured in pF, nF, or F. That’s a good reason to review your SI prefixes. And in practical capacitors you might expect a smaller A than in this example, giving even smaller capacitance. Yet capacitors of up to several F are available in small packages. In Section 16.5 we’ll explain how this is possible.