Lecture 7. Capacitors and Electric Field Energy. Last lecture review: Electrostatic potential

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Lecture 7. Capacitors and Electric Field Energy Last lecture review: Electrostatic potential V r = U r q Q

Iclicker question The figure shows cross sections through two equipotential surfaces. In both diagrams the potential difference between adjacent equipotentials is the same. Which of these two could represent the field of a point charge? A. a B. b C. Neither one 2

Iclicker question The figure shows cross sections through two equipotential surfaces. In both diagrams the potential difference between adjacent equipotentials is the same. Which of these two could represent the field of a point charge? A. a B. b C. Neither one 3

Iclicker Question A B C D 4

Iclicker Question A B C D 5

Iclicker Question Charges are fixed on a grid having the same spacing (see the Figure). Each charge has the same magnitude -Q. What are the electrostatic potential and the electric field at the location marked with an x? Assume that the reference point for V is at infinity. A. V = 0, E = 0 B. V = 0, E = k 4Q a 2 a a a C. V = k 4Q a, E = 0 a D. V = k 4Q a, E = 0 E. V = k 4Q a, E = k 4Q a 2 6

Iclicker Question Charges are fixed on a grid having the same spacing (see the Figure). Each charge has the same magnitude -Q. What are the electrostatic potential and the electric field at the location marked with an x? Assume that the reference point for V is at infinity. A. V = 0, E = 0 B. V = 0, E = k 4Q a 2 a a a C. V = k 4Q a, E = 0 a D. V = k 4Q a, E = 0 E. V = k 4Q a, E = k 4Q a 2 7

Iclicker Question The equipotential curves in a certain region of an equipotential diagram for the xy plane are parallel to the y axis and are equally spaced, with the potential increasing in the -x direction. An electric field vector in this region would point in which direction? A. +x direction B. -x direction C. +y direction D. - y direction y E. Other x 0V -1V -2V 8

Iclicker Question The equipotential curves in a certain region of an equipotential diagram for the xy plane are parallel to the y axis and are equally spaced, with the potential increasing in the -x direction. An electric field vector in this region would point in which direction? A. +x direction B. -x direction y E x dv x = dx x C. +y direction D. - y direction E. Other x 0V -1V -2V 9

Example: Dipole V = +2V V = +1V V = 0V V = 1V 10

Intersecting Equipotential Lines/Surfaces If (in 2D) an equipotential line crosses itself, then E = 0 at this point. Recall: field lines never cross! In 3D: Equipotential surfaces touch at saddle points:

Conducting Surfaces Metal Metal Metal E net? E net would drive surface current E net the only option At metallic surfaces the electric field lines are perpendicular to the surface. What can we conclude about the electric potential distribution over the surface? Conducting Surface Equipotential Surface The surface of a conductor is always an equipotential surface (in electrostatics!): when we move a test charge along the surface, no work is done (F dl). 12

Conducting Volume Since E = 0 inside, any point in the volume of a conductor is at the same potential as its surface. E=0 13

electrodes V + V - -Q +Q E Capacitors Capacitor: a system of two conducting surfaces, called electrodes. The net charge on both electrodes is typically zero. Because the surfaces are conducting, they are equipotential. electrode V = V + V = E +electrode dl is the same for any path between the electrodes. The potential difference V between the electrodes is also called the voltage across the capacitor.

Capacitance Since the charge Q on the positive electrode, and the potential difference V between the electrodes are proportional (Gauss law!), it makes sense to define the capacitance C as their ratio: C = Q V = Q electrode +electrode E dl Units: C/V = Farad (F) V = Q C Q volume of water C size (cross sectional area) of the bucket V level (height) of water in the bucket

iclicker Two conductors a and b form a capacitor. You increase the charge on a from Q to +2Q and change the charge on b from Q to 2Q, while keeping the conductors in the same positions. As a result of this change, the capacitance C of the two conductors A. becomes 4 times greater. B. becomes twice as great. C. remains the same. D. becomes 1/2 as great. E. becomes 1/4 as great. C Q V = Q electrode +electrode E dl 16

Iclicker Question The two conductors a and b form a capacitor. You increase the charge on a to +2Q and increase the charge on b to 2Q, while keeping the conductors in the same positions. As a result of this change, the capacitance C of the two conductors A. becomes 4 times great. B. becomes twice as great. C. remains the same. D. becomes 1/2 as great. E. becomes 1/4 as great. C Q V = The capacitance depends only on the geometry (shape of the electrodes, separation between them) and the medium between the electrodes. Q electrode +electrode E dl

Capacitance of Parallel-Plate Capacitor +Q E Q Recipe #1 for computing C : for a given Q, calculate V, and use the definition of capacitance. A C Q V = Q electrode +electrode E dl σ = Q A E = σ ε 0 = Q ε 0 A V = Ed = Qd ε 0 A d C = ε 0A d ε 0 9 10 12 F m Typical numbers: C 9 10 12 F m A ~ 1 m 2 d ~ 10 m = 10-5 m 1m 2 1 10 5 m 1 10 6 F 18

Iclicker Question Each diagram shows two very large parallel charged sheets with uniform charge densities and separations as shown. Which diagram corresponds to the greatest absolute value of the electric potential difference between the sheets? 19

Iclicker Question Each diagram shows two very large parallel charged sheets with uniform charge densities and separations as shown. Which diagram corresponds to the greatest absolute value of the electric potential difference between the sheets? 20

Capacitance of a coax cable (cylindrical capacitor) +Q R 2 R 1 -Q electrode V = E +electrode dl = R 2 λ dr = 2πε 0 r R 1 λ 2πε 0 ln R 2 R 1 C = Q V = λl V = 2πε 0 ln R 2 R 1 l 21

Energy Stored in a Capacitor Q i = 0 -e Q f = Q Q Initial state E = 0 d How much work does it take to charge a capacitor? d Final state E = Q ε 0 A δw = δq E q force d = δq q ε 0 A d q=q W = δq q=0 q ε 0 A d = d ε 0 A Q2 2 = Q2 2C The potential energy stored in the capacitor: U = Q2 2C = CV2 2 = QV 2 22