Physics Book Solutions

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Unformatted text preview: SET UP:! Apply Eq.(24.23) to the dashed surface in Figure 24.49: !!Q EXECUTE:! !KE + dA # encl-free !0 !! !KE + dA # KEA- since E = 0 outside the plates Qencl-free # = A- # . Q / A / AFigure 24.49 Thus KEA- # . Q/A/ A- and E # Q !0 AK !0 Qd (b) V # Ed # !0 AK !A Q Q (c) C # # # K 0 # KC0 . V (Qd/!0 AK ) d 24.50. EVALUATE:! Our result shows that K # C/C0 , which is Eq.(24.12). !A IDENTIFY:! C # 0 . C # Q / V . V # Ed . U # 1 CV 2 . 2 d SET UP:! With the battery disconnected, Q is constant. When the separation d is doubled, C is halved. ! A ! (0.16 m) 2 EXECUTE:! (a) C # 0 # 0 # 4.8 " 10!11 F d 4.7 " 10!3 m (b) Q # CV # (4.8 " 10!11 F)(12 V) # 0.58 " 10!9 C (c) E # V/d # (12 V) /(4.7 " 10!3 m) # 2550 V/m (d) U # 1 CV 2 # 1 (4.8 " 10!11 F)(12 V)2 # 3.46 " 10!9 J 2 2 (e) If the battery is disconnected, so the charge remains constant, and the plates are pulled further apart to 0.0094 m, then the calculations above can be carried out just as before, and we find:! (a) C # 2.41"10!11 F ! ! (b) Q # 0.58 "10!9 C ! Q 2 (0.58 " 10!9 C)2 # # 6.91 " 10!9 J 2C 2(2.41" 10!11 F) Q EVALUATE:! Q is unchanged. E # so E is unchanged. U doubles because C is halved. The additional stored !0 A energy comes from the work done by the force that pulled the plates apart. (c) E # 2550 V m ! ! (d) U # 24-14 Chapter 24 24.51. IDENTIFY and SET UP:! If the capacitor remains connected to the battery, the battery keeps the potential difference between the plates constant by changing the charge on the plates. !A EXECUTE:! (a) C # 0 d !12 2 (8.854 " 10 C / N + m 2 )(0.16 m)2 C# # 2.4 " 10!11 F # 24 pF 9.4 " 10!3 m (b) Remains connected to the battery says that V stays 12 V. Q # CV # (2.4 " 10!11 F)(12 V) # 2.9 " 10!10 C V 12 V # # 1.3 " 103 V/m d 9.4 " 10!3 m (d) U # 1 QV # 1 (2.9 " 10!10 C)(12.0 V) # 1.7 " 10!9 J 2 2 EVALUATE:! Increasing the separation decreases C. With V constant, this means that...
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## This document was uploaded on 03/11/2014 for the course PHYSICS 240 at University of Michigan.

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