PHY 114 Lecture Notes - Lecture 5: Electric Potential Energy, Farad, Relative Permittivity

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limeyak740

16 Feb 2018

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A capacitor is a device that stores electric potential energy by storing separated positive and negative charges. Work must be done to separate the charges. The e-field is uniform away from the plate edges. Written as an equality: q = c * v, where the proportionality constant c is called the capacitance. V = e * d = (cid:2978)(cid:2965)(cid:3116) = (cid:3436) (cid:3018)(cid:2965)(cid:3116) (cid:3002) (cid:3440) . Q = (cid:4672)(cid:2965)(cid:3116) (cid:3002)(cid:3031) (cid:4673) = c * v. 0 = 8. 85 x 10-12 (cid:3004)(cid:3118)(cid:3015) (cid:3040)(cid:3118) The unit of capacitance is the farad (f): 1 f = 1 (cid:3004)(cid:3118)(cid:3011) = 1 (cid:3004)(cid:3023) . Note: c depends only on constants and geometrical factors. Width = 0. 29 m d = 0. 01 m. C = (cid:3436)8. 85 x (cid:2869)(cid:2868)(cid:3127)(cid:3117)(cid:3118) (cid:3252)(cid:3118)(cid:3263) (cid:3288)(cid:3118) (cid:3440) ((cid:2868). (cid:2870)(cid:2870) (cid:3040) (cid:2868). (cid:2870)(cid:2877) (cid:3040)) (cid:2868). (cid:2868)(cid:2869) (cid:3040) C = 5. 65 x 10-11 (cid:3004)(cid:3023) (cid:3004)(cid:3023) is a farad, so. V = (cid:3018)(cid:3004) = (cid:3018) (cid:3031) (cid:2965)(cid:3116) (cid:3002) .

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?

beigemouse1

A capacitor is not the most efficient device for storing energy.Batteries can store more energy in much less space. For example, atypical 12 V automobile battery stores on the order of 1.00 106J.
(a) Find the capacitance necessary to store 1.00 106 J with apotential difference of 1.00 104 V across the capacitor'sterminals.
1________________

(b) Suppose that such a capacitor was made in the form of aparallel-plate capacitor with a vacuum between the plates and anelectric field no greater than 7.00 106 V/m. What is the minimumarea of the plates?
2 ________________

And.........

The plates of a parallel-plate capacitor each have an area of 0.1m2 and are separated by a 1 mm thick layer of paper. The capacitoris connected to a 11 V battery. (The dielectric constant for paperis 4.)
(a) Find the capacitance.
1

(b) Find the charge stored.
2

(c) Find the electric field between the plates.
3

Two parallel plates are charged with +Q and -Q respectively as shown in the figure, where Q = 26 nC. The area of each plate is A = 0.019 m2 The distance between them is d = 6.4. cm. The plates are in the air.

Part a. With the information given, write the equation of the capacitance in terms of ε0, A, d.

Part b. Solve for the numerical value of C in pF

Part c. With the information given, express the potential difference across the capacitor in terms of Q and C.

Part d. Solve for the numerical value of ΔV in V.

Part e. If the charges on the plates are increased to 2Q, what is the value of the capacitance C in pF?

Part f. If A is then increased to 2A, d is decreased to d/4, what is the value of the capacitance C in pF?

Part g. What is the new value of potential difference ΔV in V with the original charge Q, given the values for A and d from part (f)?

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PHY 114 Lecture Notes - Lecture 5: Electric Potential Energy, Farad, Relative Permittivity

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