Chapter20[1]

Chapter20[1] - Chapter 20 Review gravitational/spring...

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Chapter 20 Review gravitational/spring potential energy and conservative forces ( Δ U only depends on the beginning/ending points) from last semester. Example: If a rock is lifted in a gravitational field, its PE will increase. Why? Δ U = -W ˆˆ W F d ( mgy) (dy) mgd   Ask yourself: Will it hurt more to get “beaned” by the rock if it is at a higher altitude? It has more “ potential” to whack you! The electric force is also conservative ! Consider the analogous situation for a charge: + q 0 E Just the same as with the rock, the test charge “wants” to follow the arrow of the E -field if we were to let go of it. So, if the charge is raised, it should INCREASE the potential. W = F · d = q 0 E · d = q 0 E (- y ) · d (+ y ) = - q 0 E d Δ U = - W = q 0 Ed So, what would happen if the rock is lowered? This should DECREASE the potential! W Fd qEd qE (y )d ) qE d Δ UWq E d      00 0 0 Let Ed define a new quantity – the electric POTENTIAL. We set V = 0 at any place we want, often at infinity, but not always. V
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00 UW V qq   Joule/Coulomb = Volt or 1 V = 1 J/C We can rearrange: 1 J = (1 V)(1 C) This is too big to be useful. 1 electron volt (eV) is the energy needed to push 1 electron (1.6 x 10 -19 C) through a potential difference of 1 V. 1 eV = 1.6 x 10 -19 J. EXAMPLE: Find Δ U when 12 q 2.2 C or q 1.10 C  moves from point A to point B if Δ V AB = V B(end) – V A(start) = 24.0 V 5 11 5 22 U V q U (q )( V) (2.2 C)(24.0V) 5.28x10 J U (q )( V) ( 1.1 C)(24.0V) 2.64x10 J Connection between electric field and potential: When a positive charge moves along the line of E, the potential goes down (rock falling in a gravitational field). 0 0 UW( q E s ) VE q V EV / m s   s x E Note: 1 N/C = 1 V/m q 1 q 2
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1. V decreases as we move in the direction of E. 2. Δ V is negative when E and Δ s are both positive. 3. Remember that E originates on (+) and terminates on (-), so V goes down as we move to regions that are more negative ( for the positive test charge ). EXAMPLE: V = 12 V Δ s = 0.75 cm Find E.
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BA V( V V ) ( 0 V 1 2 V ) (a) E 1600V / m. This should be s s (0 0.0075m) negative, because the electric field points in the minus y-direction. (b) q 6.24 C moves from positive plate to negative plate. Fin     d U.
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This note was uploaded on 10/13/2011 for the course PHY 102 taught by Professor Alexandrakis during the Fall '06 term at FIU.

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Chapter20[1] - Chapter 20 Review gravitational/spring...

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