26. Capacitors & Dielectrics

To get an idea how big a farad is, suppose you want to make a 1-F air-filled parallel-plate capacitor for a circuit you are building. To make it a reasonable size, suppose you limit the plate area to 1.0 cm². What would the gap have to be between the plates? Is this practically achievable?

In lightning storms, the potential difference between the Earth and the bottom of the thunderclouds can be as high as 35,000,000 V. The bottoms of thunderclouds are typically 1500 m above the Earth, and can have an area of 110 km². Modeling the Earth–cloud system as a huge capacitor, calculate the capacitance of the Earth–cloud system,

A parallel plate capacitor consists of two plates of area $A$ separated by a distance $d$ and filled with a dielectric of permittivity $\epsilon$. Which of the following gives the correct expression for its capacitance?

A capacitor is made from two 1.1-cm-diameter coins separated by a 0.10-mm-thick piece of paper (K = 3.7). A 12-V battery is connected to the capacitor. How much charge is on each coin?

A $5.00$-$\mu$F parallel-plate capacitor is connected to a $12.0$ V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates.

(a) A voltmeter is connected across the two plates without discharging them. What does it read?

(b) What would the voltmeter read if (i) the plate separation were doubled; (ii) the radius of each plate were doubled but their separation was unchanged?

Two circular plates of radius 2cm are brought together so their separation is 5mm. What is the capacitance of these plates?

A parallel-plate air capacitor is to store charge of magnitude $240.0$ pC on each plate when the potential difference between the plates is $42.0$ V.

(a) If the area of each plate is $6.80$ cm², what is the separation between the plates?

(b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude $240.0$ pC on each plate?

A student wearing shoes with thin insulating soles is standing on a grounded metal floor when he puts his hand on the metal case of a high voltage supply. The voltage inside, 6.3 mm from his hand, is 25,000 V. The student’s hand and the voltage supply form a capacitor, considering the student as a conductor. Another capacitor is between the floor and the student (his feet). Using reasonable numbers for hand and foot areas, estimate the student’s voltage relative to the floor. Assume vinyl-soled shoes 1 cm thick.

Small distances can be measured using a capacitor whose plate separation 𝓍 is variable. Consider an air-filled parallel-plate capacitor with fixed plate area A = 25 mm² separated by a variable distance 𝓍. Assume this capacitor is attached to a capacitance-measuring instrument which can measure capacitance C in the range 1.0 pF to 1000.0 pF with an accuracy of ∆C = 0.1 pF. Define ∆𝓍 to be the accuracy (magnitude) to which 𝓍 can be determined, and determine a formula for ∆𝓍.

A 3 F capacitor is given a potential difference across its plates of 10 V. What is the charge built up on its plates? If the source of the potential difference across the plates is removed, but the plates maintain their charge, what is the new potential difference across the capacitor if the distance between the plates is doubled?

(I) What is the capacitance per unit length (F/m) of a coaxial cable whose inner conductor has a 1.0-mm diameter and the outer cylindrical sheath has a 5.0-mm diameter? Assume the space between is filled with air.

The plates of a parallel-plate capacitor are $2.50$ mm apart, and each carries a charge of magnitude $80.0$ nC. The plates are in vacuum. The electric field between the plates has a magnitude of $4.00\times {10}^6$ V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?