Physics 212 Lecture 6 Electric Potential Physics 212

Physics 212 Lecture 6 Electric Potential Physics 212 Lecture 6, Slide 1

BIG IDEA • Last time we defined the electric potential energy of charge q in an electric field: • The only mention of the particle was through its charge q. • We can obtain a new quantity, the electric potential, which is a PROPERTY OF THE SPACE, as the potential energy per unit charge. V in Volts = Joules/Coulomb • Note the similarity to the definition of another quantity which is also a PROPERTY OF THE SPACE, the electric field. 40 Physics 212 Lecture 6, Slide 2

Electric Potential from E field • Consider the three points A, B, and C located in a region of constant electric field as shown. D Dx • What is the sign of DVAC = VC - VA ? (A) DVAC < 0 (B) DVAC = 0 (C) DVAC > 0 • Remember the definition: • Choose a path (any will do!) 40 Physics 212 Lecture 6, Slide 3

Checkpoint 2 A B C D • Remember the definition V is constant !! 08 Physics 212 Lecture 6, Slide 4

E from V • We obtain the potential by integrating the electric field: • So, we can obtain the electric field by differentiating the potential. • In Cartesian coordinates: 40 Physics 212 Lecture 6, Slide 5

Checkpoint 1 a • How do we get E from V? ? Look at slopes !!! 08 Physics 212 Lecture 6, Slide 6

Checkpoint 1 b A B C D • How do we get E from V? ? Look at slopes !!! 08 Physics 212 Lecture 6, Slide 7

Equipotentials • Equipotentials are the locus of points having the same potential. Equipotentials produced by a point charge Equipotentials are ALWAYS perpendicular to the electric field lines The SPACING of the equipotentials indicates The STRENGTH of the electric field 40 Physics 212 Lecture 6, Slide 8

Checkpoint 3 a A B C D 08 Physics 212 Lecture 6, Slide 9

Checkpoint 3 b A B C D 08 Physics 212 Lecture 6, Slide 10

HINT E - FIELD LINES A B C D EQUIPOTENTIALS • What is work done by E field to move negative charge from A to C ? (A) WAC < 0 (B) WAC = 0 (C) WAC > 0 A and C are on the same equipotential 08 Equipotentials are perpendicular to the E field: No work is done along an equipotential. WAC = 0 Physics 212 Lecture 6, Slide 11

Checkpoint 3 b Again? A B C D • A and C are on the same equipotential • B and D are on the same equipotential • Therefore the potential difference between A and B is the SAME as the potential between C and D 08 Physics 212 Lecture 6, Slide 12

Checkpoint 3 c A B C D 08 Physics 212 Lecture 6, Slide 13

cross-section a 4 a 3 +Q a 2 a 1 +q metal Point charge q at center of concentric conducting spherical shells of radii a 1, a 2, a 3, and a 4. The inner shell is uncharged, but the outer shell carries charge Q. What is V(r) as a function of r? metal - Charges q and Q will create an E field throughout space – – 40 Spherical symmetry: Use Gauss’ Law to calculate E everywhere Integrate E to get V Physics 212 Lecture 6, Slide 14

a 4 a 3 r > a 4 : What is E(r)? +Q a 2 a 1 (A) 0 (B) (C) +q metal r (D) (E) metal Why? Gauss’ law: Physics 212 Lecture 6, Slide 15

a 4 a 3 +Q a 2 a 1 (A) 0 +q metal a 3 < r < a 4 : What is E(r)? r (B) (D) (C) (E) metal Applying Gauss’ law, what is Qenclosed for red sphere shown? (A) q (B) –q (C) 0 How is this possible? ? ? -q must be induced at r=a 3 surface charge at r=a 4 surface = Q+q Physics 212 Lecture 6, Slide 16

a 4 a 3 +Q a 2 a 1 +q metal Physics 212 Lecture 6, Slide 17

a 4 a 3 +Q First find V(r) for r a 4 a 2 a 1 +q metal 0 For r a 4 Just like a point charge ! So…

a 4 a 3 +Q How about V(r) for a 2 a 1 ? +q metal 0 V( r ) constant inside a conductor !

a 4 a 3 +Q V(r) for a 2 a 1 +q metal 0 ?

a 4 a 3 +Q ? V(r) for a 2 a 1 +q metal 0 0

a 4 a 3 +Q V(r) for a 2 a 1 ? +q metal 0 0

Summary V 0 a 1 a 2 a 3 a 4 r