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Unformatted text preview: U / U 0 2 2 K# 24.46. U 1.85 " 10!5 J 0 2.32 " 10!5 J # # 2.25 1.85 " 10!5 J U0 EVALUATE:! K increases the capacitance and then from U # 1 CV 2 , with V constant an increase in C gives an 2 increase in U. IDENTIFY:! C # KC0 . C # Q / V . V # Ed . SET UP:! Since the capacitor remains connected to the battery the potential between the plates of the capacitor doesn't change. EXECUTE:! (a) The capacitance changes by a factor of K when the dielectric is inserted. Since V is unchanged (the C Q 45.0 pC # K # 1.80 . battery is still connected), after # after # Cbefore Qbefore 25.0 pC (b) The area of the plates is , r 2 # , (0.0300 m) 2 # 2.827 " 10!3 m 2 and the separation between them is thus !0 A (8.85 " 10!12 C2 N + m 2 )(2.827 " 10!3 m 2 ) !A Q # # 2.00 " 10!3 m . Before the dielectric is inserted, C # 0 # 12.5 " 10!12 F C d V Qd (25.0 " 10!12 C)(2.00 " 10!3 m) and V # # # 2.00 V . The battery remains connected, so the potential !0 A (8.85 " 10!12 C2 /N + m 2 )(2.827 " 10!3 m 2 ) difference is unchanged after the dielectric is inserted. Q 25.0 " 10!12 C (c) Before the dielectric is inserted, E # # # 1000 N/C !0 A (8.85 " 10!12 C2 /N + m 2 )(2.827 " 10!3 m 2 ) Again, since the voltage is unchanged after the dielectric is inserted, the electric field is also unchanged. 2.00 V V EVALUATE:! E # # # 1000 N/C , whether or not the dielectric is present. This agrees with the result d 2.00 " 10!3 m in part (c). The electric field has this value at any point between the plates. We need d to calculate E because V is the potential difference between points separated by distance d. IDENTIFY:! C # KC0 . U # 1 CV 2 . 2 d# 24.47. SET UP:! C0 # 12.5 $ F is the value of the capacitance without the dielectric present. EXECUTE:! (a) With the dielectric, C # (3.75)(12.5 $ F) # 46.9 $ F . before: U # 1 C0V 2 # 1 (12.5 " 10!6 F)(24.0 V)2 # 3.60 mJ 2 2 24.48. after: U # 1 CV 2 # 1 (46.9 " 10!6 F)(24.0 V)2...
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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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