# lecture 10 - Announcement:  Clickers:

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Unformatted text preview: Announcement:  Clickers:   75% of ques6ons answered per session: 3 points  Each ques6on answered correctly: 1 more point  In‐class Quiz: Tuesday, 30 minutes, 3 ques6ons Start 9 am Sharp  Handing it in worth 2 clicker points  Each ques6on worth 2 clicker points   r q  Q  Coulomb’s Law  Electric Field  F= 1 Qq ˆ r 2 4πε 0 r F = qE E (r ) = 1Q ˆ r 2 4πε 0 r Gauss’s Law  Flux over a closed surface  Φtotal Φtotal Qenclosed = ε0 ˆ = ∫ E ⋅ ndS Electric Poten6al Energy on charge q:    U(q)   (J)           Work done by electric ﬁeld is independent of path,                                                      but dependent on charge  Electric Poten6al: V = U/q  (V)   Example : Parallel Plate Capacitor  σ= 1 µC/m2  +σ  a  1 m   b   1 µC  ‐σ  1 × 10 −6 C / m Electric ﬁeld between parallel plates = σ/ε0=     = 1.1 × 10 5 N / C −12 2 8.85 × 10 ⋅ C / Nm Work done by my hand = qEd =  (1 × 10 −6 C ) × (1.1 × 10 5 N ) × (1m ) = 0.11Nm = 0.11J C σ= 1µC/m2  +σ  a  1 µC  1 m   b   ‐σ  Poten6al Energy at posi6on a  U(a)‐U(b) =0.11 J  Poten6al   (U(a)‐U(b))/q = V(a)‐V(b) = 0.11 J/1µC=1.1×105 V   Problem 1: When a nega6ve charge is  released and moves along an electric ﬁeld  line, it moves to a posi6on of  a.  b. c.  d.  e.  lower poten6al and lower  poten6al energy.   lower poten6al and higher  poten6al energy.  higher poten6al and lower  poten6al energy.  higher poten6al and  higher poten6al energy.   decreasing magnitude of  the electric ﬁeld.  Example: Q sta6onary: how much energy does it take to bring q to the below posi6on?  R Q  F= 1 Qq ˆ r 2 4πε 0 r At distance r  Work done by hand = Required Energy    q  How much energy does it take to move it an inﬁnitesimal distance in at distance r?  ˆ − drr q  Q  1 Qq 1 Qq ˆ ⋅ − drr = − ˆ Work = Force × Dis tan ce = r dr 2 2 4πε 0 r 4πε 0 r Worktotal 1 Qq 1 Qq =∫ − dr = 2 ∞ 4πε 0 r 4πε 0 R R ˆ x Q  1Q V (x) = 4πε 0 x E= 1Q dV ˆ ˆ x=− x 4πε 0 x 2 dx X axis  r q  Q  Poten6al Energy at R :   Poten6al at R:   E= 1 Qq 4πε 0 r 1Q V (r ) = 4πε 0 r 1 Qq dV ˆ ˆ r=− r 2 4πε 0 r dr In general,       Gradient Discussion: Board  Problem 2: The electric ﬁeld in a region of space is given by   Ex = (3.0x) N/C, Ey = Ez = 0, where x is in m. Points A and B are  on the x axis at xA = 3.0 m and xB = 5.0 m. Determine the  poten6al diﬀerence VB – VA.  a.  b.  e.  d. e. –24 V  +24 V  –18 V   +30 V   –6.0 V  Problem 3: Points A [at (3, 6) m] and B [at (8, –3) m] are in a  region where the electric ﬁeld is uniform and given by E = 12i  N/C. What is the electric poten6al diﬀerence VA – VB ?  a. b. c. d. e.  +60 V   –60 V   +80 V   –80 V   +50 V  Problem 4: Points A [at (2, 3) m] and B [at (5, 7) m] are in a  region where the electric ﬁeld is uniform and given by E = (4i +  3j) N/C. What is the poten6al diﬀerence VA – VB ?  a. b. c.  d.  e.   33 V   27 V  30 V  24 V  11 V  Example: Conduc6ng Sphere with total charge Q  Radius : R  P  r Example: Uniformly Charged Sphere with total charge Q  Radius : R  P  r What about a non‐tradi6onal objects?  Conduc6ng irregular shaped object  Radius b  Radius a  ...
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## This note was uploaded on 07/16/2011 for the course PHY 2049 taught by Professor Saha during the Fall '08 term at University of Central Florida.

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