Two identical capacitors are joined in parallel, charged to a potential V, separated and then connected in series, the positive plate of one is connected to the negative of the other. Which of the following is true?
(a) The charges on the free plated connected to-gether are destroyed.
(b) The energy stored in ths system increases.
(c) The potential difference between the free plates is 2V.
(d) The potential difference remains constant.
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If two identical capacitors are joined in parallel, charged to a potential V, separated and then connected in series then (d) The potential difference remains constant.
- Since, the free plate charge is isolated it can't be altered.
- Since the charge appears on free plates, the charge present on the internal plates can not be changed.
- Therefore, E inside conductors are zero so q must be 0. So, there is the same negative charge on the other plate
- Q1 = CV1 and Q2 = CV2 Applying the charge conservation
CV1 + CV2 = Q1+Q2
CV1 + CV2 = 2CV
Þ V1+V2 = 2V
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