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Unformatted text preview: s of the capacitor voltage. SET UP:! Le the applied voltage be V. Let each capacitor have capacitance C. U # 1 CV 2 for a single capacitor with 2 voltage V. EXECUTE:! (a) series Voltage across each capacitor is V /2. The total energy stored is Us # 2 . 1 C[V/2]2 / # 1 CV 2 2 4 parallel Voltage across each capacitor is V. The total energy stored is Up # 2 . 1 CV 2 / # CV 2 2 Up # 4Us (b) Q # CV for a single capacitor with voltage V. Qs # 2 . C[V/2]/ # CV ; Qp # 2(CV ) # 2CV ; Q p # 2Qs (c) E # V/d for a capacitor with voltage V. Es # V/2d ; Ep # V/d ; Ep # 2 Es 24.38. EVALUATE:! The parallel combination stores more energy and more charge since the voltage for each capacitor is larger for parallel. More energy stored and larger voltage for parallel means larger electric field in the parallel case. IDENTIFY:! V # Ed and C # Q / V . With the dielectric present, C # KC0 . SET UP:! V # Ed holds both with and without the dielectric. EXECUTE:! (a) V # Ed # (3.00 " 104 V/m)(1.50 " 10!3 m) # 45.0 V . Q # C0V # (5.00 " 10!12 F)(45.0 V) # 2.25 " 10!10 C . (b) With the dielectric, C # KC0 # (2.70)(5.00 pF) # 13.5 pF . V is still 45.0 V, so Q # CV # (13.5 " 10!12 F)(45.0 V) # 6.08 " 10!10 C . EVALUATE:! The presence of the dielectric increases the amount of charge that can be stored for a given potential difference and electric field between the plates. Q increases by a factor of K. Capacitance and Dielectrics 24.39. 24-11 IDENTIFY and SET UP:! Q is constant so we can apply Eq.(24.14). The charge density on each surface of the dielectric is given by Eq.(24.16). E E 3.20 " 105 V/m EXECUTE:! E # 0 so K # 0 # # 1.28 K E 2.50 " 105 V/m (a) = i # = (1 ! 1/ K ) = # !0 E0 # (8.854 " 10!12 C 2 /N + m 2 )(3.20 " 105 N/C) # 2.833 " 10!6 C/m 2 24.40. = i # (2.833 " 10!6 C/m 2 )(1 ! 1/1.28) # 6.20 " 10!7 C/m 2 (b) As calculated above, K # 1.28. EVALUATE:! The surface charges on the dielectric produce an electric field that part...
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