Electrical Charge

Introduction

Mr. Volta’s pile was prepared, built of round pieces of zinc and copper, alternating with pieces of humid flannel with water impregnated with ammoniac salt; a silver wire, divided in several pieces like a chain, was attached to the pile. The last segment of the chain passed through a glass tube, and from its exterior termination there stretched a pure silver knob attached to the chain. After this was done, he [the experimenter, Giandomenico Romagnosi] took an ordinary magnetic needle, in the form of a nautical compass,boxed in a square wood plank; and lifting up the crystal that enclosed it, he put it over a glass insulator, near the aforementioned pile.28Then, taking the silver chain and holding it by the above-mentioned glass tube, he applied its end or knob to the magnetic needle; and keeping the contact for a few seconds, he made the needle deflect from the direction of the poles by a few degrees. Taking off the silver chain,the needle remained steady in the divergent direction that was given to it. He again applied the same chain, and made the needle diverge even more from the polar direction, and the needle always remained in the same place in which it was left, in such a way that the polarity remained completely damped.

— Giandomenica Romagnosi, Description of His Experiments in Magnetism, May, 1802

Outline

Storing Charge
- Capacitors
- Dielectrics
Practice with the Concepts
Discussion Points

Storing Charge

Electric Potential, Electric Energy, Capacitance

Last chapter, we talked about collecting excess charge on objects, and how to create that excess by induction or direct conduction. Having excess charge around was very important during the early investigations into electrical phenomena, to supply both static charge and current. Around 1746, the Dutch physicist Pieter van Musschenbroek invented a capacitor known as the Leyden jar, a glass or ceramic cup lined inside and out with a conducting metal. For the next 150 years, Leyden jars or variations on them served as capacitors until the invention of radios precipitated a need for more precise control of the charge storage capacity. Now, being able to store and discharge electricity at will is a fundamental capability required by most electronic equipment.

Capacitors

Modern capacitors or condensors consist of two conducting materials, often flat plates, placed near each other but not touching. The space between is the plates is called the dielectric medium, and it can be air, or vacuum, or some other non-conducting medium. One of the materials is charged by conduction, the other is not (but becomes charged in an equal amount by induction), creating a potential difference and an electric field between them. This potential is a function of the capacitance or the material's ability to hold charge C, and the actual charge on one of the plates: $Q = CV where C = ϵ_{0} \frac{A}{d}$

Any one of the three versions of this relationship can be very useful. $Q = CV and C = \frac{Q}{V} and V = \frac{Q}{C}$ Capacitance is measured in farads, or volts/coulomb (which should be obvious from the C = V/Q relationship).

Capacitance is constant for a given setup. If the surface area for the plates is A and the distance between the plates is d, then the capacitance: $C = ϵ_{0} \frac{A}{d}$ You will recall that ε₀ is the permittivity of free space.

Actually getting charge into the capacitor is a bit tricky. Since V = Q/C and V is itself an indication of the work involved to move charge, $W = Fd = qEd = qV$ as the charge already stored in the capacitor grows, the work necessary to move the next charge bit into the capacitor increases. We will need to use the average voltage $V_{avg} = \frac{V_{b} - V_{a}}{2}$ in order to determine the total work necessary to charge a capacitor. If we start with V_a = 0, however (a completely uncharged capacitor), the work becomes $W = Q_{tot} V_{avg} = Q_{tot} (\frac{V_{b}}{2})$ This is also the potential (electrical) energy U stored in the capacitor. By manipulating our favorite formulae for V, Q, and C, we get the relationships

PE = \frac{1}{2} QV = \frac{1}{2} (CV) V = \frac{1}{2} C V^{2} = \frac{1}{2} Q (\frac{Q}{C}) = \frac{1}{2} \frac{Q^{2}}{C}

We use whichever one is appropriate for the given situation.

Dielectrics

Sometimes the space between plates in a capacity is filled with material which can hold the plates apart and hold more charge than air alone. The equation for capacitance in a capacitor using a dielectric is

C = {K ϵ}_{0} \frac{A}{d}

Since the dielectric constant of any material K > 1 (vacuum is defined as K = 1.000) the capacitance C of a system using a dielectric is greater than the capacitance without it. Using a dielectric therefore increases the amount of charge which can be stored in a given capacitor.

Practice with the Concepts

Discussion Points

When a battery is connected to a capacitor, why do the two plates acquire charges of the same magnitude? Will this be true if the two plates are different sizes or shapes?
The parallel plates of an isolated capacitor carry opposite charges, Q. If the separation of the plates is increased, is a force required to do so? Is the potential difference changed? What happens to the work done in the pulling process?

© 2005 - 2025 This course is offered through Scholars Online, a non-profit organization supporting classical Christian education through online courses. Permission to copy course content (lessons and labs) for personal study is granted to students currently or formerly enrolled in the course through Scholars Online. Reproduction for any other purpose, without the express written consent of the author, is prohibited.

Physics

Chapter 17: 7-11