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Unformatted text preview: Q 1 > Q 2 Q 1 = Q 2 Q 1 < Q 2 (ii) If the two capacitors are in series, compare the potential difference over each capacitor: V 1 > V 2 V 1 = V 2 V 1 < V 2 (iii) If the two capacitors are in parallel, compare the charge on each capacitor Q 1 > Q 2 Q 1 = Q 2 Q 1 < Q 2 (iv) If the two capacitors are in parallel, compare the potential difference over each capacitor: V 1 > V 2 V 1 = V 2 V 1 < V 2 Activity 3: The figure shows a parallel plate capacitor with a capacitance C 1 = 140 pF which has been charged by a battery to a potential difference of 80 V. The battery is then disconnected. Now a second capacitor C 2 , which is initially uncharged, is connected in parallel to C 1 . This causes the potential difference over C 1 to drop to 50 V. Use this information to find the capacitance C 2 . (Hints: which quantity is conserved here? And what is the equivalent capacitance in the combination?)...
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