PHYSICS 7D Lecture Notes - Lecture 10: Farad, Electric Field, Electrical Network

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PHYSICS 7D Lecture Notes - Lecture 8: Faraday Cage, Equipotential, Electric Field

PHYSICS 7D Lecture Notes - Lecture 10: Farad, Electric Field, Electrical Network

PHYSICS 7D Lecture Notes - Lecture 11: Valence And Conduction Bands, Charge Carrier, Electric Current

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Potential difference v between plates imposed with battery. +q will be stored on one conductor, -q on the other conductor. A battery has a fixed potential difference between the terminals. Charges on the conductors establishes an e field between them. The difference in potential between the plates s equal the potential difference produced by the battery. Q will be a function of v and the geometry of the capacitor. The ratio q/v is a constant for the given geometry. Q/v is a constant in a capacitor. Capacitance is the ability of a capacitor (two separated conductors) to store charge. Si unit of capacitance is farad (f) Capacitance depends only on geometry, not q or v. Assume a charge q is on each of the conductors. No charge flows in the space between the conductors--charge gets deposited on them by some external source. Calculate electric field between the conductors using gauss"s law.

Related Questions

To understand the meaning of capacitance and ways of calculating capacitance.

When a positive charge q is placed on a conductor that is insulated from ground, an electric field emanates from the conductor to ground, and the conductor will have a nonzero potential V relative to ground. If more charge is placed on the conductor, this voltage will increase proportionately. The ratio of charge to voltage is called the capacitance C of this conductor: C=q/V.

Capacitance is one of the central concepts in electrostatics, and specially constructed devices called capacitors are essential elements of electronic circuits. In a capacitor, a second conducting surface is placed near the first (they are often called electrodes), and the relevant voltage is the voltage between these two electrodes.

This tutorial is designed to help you understand capacitance by assisting you in calculating the capacitance of a parallel-plate capacitor, which consists of two plates each of area A separated by a small distanced with air or vacuum in between. In figuring out the capacitance of this configuration of conductors, it is important to keep in mind that the voltage difference is the line integral of the electric field between the plates.

Assume that charge -q is placed on the top plate, and +q is placed on the bottom plate. What is the magnitude of the electric field E between the plates?
What is the voltage V between the plates of the capacitor?

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Suppose a parallel plate capacitor (with capacitance C0) is fully charged (to a value Q0) by a battery. The battery (which supplies a potential difference of V0) stays connected to the capacitor. If the plates of the capacitor are then moved closer together (the separation distance d b/w the plates is halved), describe QUANTITATIVELY what happens to:

a). the capacitance of the capacitors

b). the potential difference b/w the plates

c). the energy stored in capacitor

d). the charge on the plates.

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PHYSICS 7D Lecture Notes - Lecture 10: Farad, Electric Field, Electrical Network

PHYSICS 7D Lecture Notes - Lecture 8: Faraday Cage, Equipotential, Electric Field

PHYSICS 7D Lecture Notes - Lecture 10: Farad, Electric Field, Electrical Network

PHYSICS 7D Lecture Notes - Lecture 11: Valence And Conduction Bands, Charge Carrier, Electric Current

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