16 - 16.P.37 V along different paths in a capacitor (a) For...

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16.P.37 Δ V along different paths in a capacitor (a) For the path 1, A - B - C shown in the diagram, going from A to B we’re go- ing directly opposite to the electric field (magnitude E ) so the potential increas- es by an amount Es 1 . From B to C the electric field is perpendicular to the path, so is zero. Therefore Δ V = V C – V A = Es 1 For path 2, A - C, we have since all the dx ’s are negative, but E x is positive (for x axis pointing to the right). For path 3, A - D - B - C , along A - D there is no contribution (field perpendicular to path); along D - B the potential difference is Es 1 (same as A - C ), and along B - C there is no contribution (field perpendicular to path). So again the potential difference is Es 1 . (b) (c) For A - B - C - D , along C - D , Δ V is the opposite of A - B , so the round trip gives zero. For A - B - A , again there is cancellation: A - B is + Es 1 , and B - A is – Es 1 , so the round trip gives zero. A B CD +Q –Q s 1 s 2 s R >> s E El d d i f E x x d i f Es 1 () 1 === V C V A QA ε 0 ---------------- s 1 43 6 × 10 C π 4 m 2 ------------------------------ 0.7 3 × 10 m 9 12 × 10 C 2 N·m 2 --------------- ⎝⎠ ⎛⎞ ---------------------------------------- 67 volts == =
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16.P.40 Δ V difference near a dipole Fields not drawn to scale on this diagram: A B C D s d l v v E v E v E v E d l v d l v d l v (a) Everywhere on the path from A to B the field is perpendicular to the path, so we have Δ V =− v E d v l d 0 = 0 . (b) Δ V = V C V D v E D C d v l = Edl D C since v E // d v l 1 4 πε 0 b a 2 sq x 3 d l and since d v l d v x =+ 1 4 0 b a 2 sq x 3 dx = 2 sq 4 0 dx x 3 b a = 2 sq 4 0 1 2 1 x 2 b a Δ V sq 4 0 1 a 2 1 b 2 b > a so 1 b 2 < 1 a 2 and therefore 1 a 2 1 b 2 ⎟ > 0 so Δ V < 0, as it should be, since we are traveling with the electric field. It is very easy to make a sign error in this integral; determine the sign on physical grounds instead of trying to get it right in the math (i.e. it's ok to fix the sign on physical grounds!) Alternatively, let’s find the potential at a distance x from the center of this dipole: Therefore we have as before the following result: (c) Δ U = q Δ V e () sq 4 0 1 a 2 1 b 2 Δ U > 0 as it should : an external force must do work on the electron to push it in the direction of the electric field. ⎡− + + = + + = 2 0 0 0 4 1 ) 2 / )( 2 / ( ) 2 / ( ) 2 / ( 4 2 / 2 / 4 1 x qs s x s x s x s x q s x q s x q V = Δ 2 2 0 1 1 4 b a sq V
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16.P.41 Field and Δ V in a capacitor (a)
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16.P.43 Δ V near a charged spherical shell A C B D +Q r 1 + + + + + + + + + + + + E = 0 inside dl dl E E E E dl Path 1, A - B - C From A to the surface, we go against the field and the potential rises: V right surface V A = Q 4 πε 0 1 R 1 r 1 > 0 , where R is the radius of the sphere All along the path from the right surface to the left surface, the field is zero, so Δ V = 0.
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16 - 16.P.37 V along different paths in a capacitor (a) For...

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