hw7 - Physics 21 Fall, 2005 Solution to HW-7 24 P 59 Two...

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Physics 21 Fall, 2005 Solution to HW-7 24 P59 Two identical capacitors are connected in parallel, and each acquires a charge Q 0 when connected to a source of voltage V 0 . The voltage source is disconnected and then a dielectric ( K =2 . 5) is inserted to Fll the space between the plates of one of the capacitors. Determine (a) the charge now on each capacitor, and (b) the voltage now across each capacitor. +Q 0 +Q 0 2Q 0 - QQ C 0 C 0 C 0 KC 0 (a) (b) ±or panel (a), the two capacitors are the same and Q 0 = C 0 V 0 . ±or panel (b), the charge on the two upper plates will rearrange so that the voltages across the two capacitors ( C 0 on the left and KC 0 on the right) are equal. Since the capacitors are in parallel, the total charge on the two positive plates does not change. We can Fnd the charge Q that ends up on the modiFed capacitor by using the equality of voltages: 2 Q 0 Q C 0 = Q 0 Solving this equation for Q gives Q Q 0 K K +1 and 2 Q 0 Q Q 0 1 K . The voltage across both capacitors is V = Q 0 = Q 0 C 0 2 K = V 0 2 K .
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