# HW6solutions - Physics 2213 HW #6 Solutions Spring 2009...

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Physics 2213 HW #6 – Solutions Spring 2009 24.14. Let C 1 15 pF , C 2 9.0 pF and C 3 11 pF . The capacitors between b and c , that is C 2 and C 3 , are in parallel. The equivalent capacitance of this combination is C 23 C 2 C 3 20 pF . The combination C 23 is in series with the C 1 (15 pF) capacitor, so the total effective capacitance is given by 1 C 123 1 C 1 1 C 23 . Solving for C 123 , C 123 C 1 C 23 C 1 C 23 (15 pF)(20 pF) 15 pF 20 pF 8.6 pF Notice that for capacitors in parallel the equivalent capacitance is larger than any of the individual capacitors. For capacitors in series the equivalent capacitance is smaller than any of the individual capacitors. 24.66. If there is charge +Q on the top plate and –Q on the bottom plate, then the metal slab, being a conductor, will acquire an induced charge of –Q on its top and +Q on its bottom. This situation is analogous to having two capacitors C 1 in series, each with separation 1 2 ( d a ). Recall that for capacitors in series 1 C eq 1 C 1 1 C 2 . In our case, C 1 =C 2 . The capacitance of a parallel plate capacitor is 0 A C s , where s is the plate separation. For C 1 , s= ( d-a )/2. For C 0 (the capacitor without the metal slab), s=d . (a) C 1 C 1 1 C 1 1 1 2 C 1 1 2 0 A ( d a ) 2 0 A d a (b) C 0 A d a 0 A d d d a C 0 d d a (c) As a 0 , C C 0 . The metal slab has no effect if it is very thin. And as a d , C  . V Q / C . V Ey is the potential difference between two points separated by a distance y parallel to a uniform electric field. When the distance is very small, it takes a very large field and hence a large Q on the plates for a given potential difference. Since Q CV this corresponds to a very large C . #1 ["Build-a-Cap"] (a) A = (0.17 m)(0.24 m) = 0.041 m 2 , and C =  o A d = (1.5)(8.85 x 10 -12 F/m) 0.041 m 2 1.0 x 10 -4 m = 5.4 x 10 -9 F = 5.4 nF V max = E max d = (7.0 x 10 6 V/m)(1.0 x 10 -4 m) = 700 V (b) We can still use the parallel plate capacitor approximation since the spacing between the foil sheets remains small compared with the radius of the rolled up capacitor. But now both sides of each foil sheet face a side of the other foil sheet with either one or two sheets of paper in-between. This arrangement is equivalent to two capacitors connected in parallel — one with the capacitance calculated in part (a) and the other with twice the thickness of paper in-between (two sheets) so that it has ½ of the capacitance in (a).

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## This note was uploaded on 04/14/2009 for the course PHYS 2213 taught by Professor Perelstein,m during the Spring '07 term at Cornell University (Engineering School).

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HW6solutions - Physics 2213 HW #6 Solutions Spring 2009...

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