Two identical capacitors are joined in parallel, charged to a potential V and then separated and connected in series, i.e., negative plate of first is connected to positive plate of second :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → the potential differences between the free plates becomes 4 V
Note: The given answer is as per the official NTA Answer Key. However, there seems to be a mistake in the key: two capacitors each at potential V in series give 2V, so the correct answer should be Option (3), as explained below.
Working as per the NTA key
Charged in parallel to potential \( V \), the two capacitors (each \( C \)) hold a total charge \( 2CV \). Placing this on the series combination \( C/2 \) gives
\( \displaystyle V' = \frac{Q_{\mathrm{total}}}{C/2} = \frac{2CV}{C/2} = 4V \)
This assumes the charges of the two capacitors add on the series combination. They do not: each capacitor keeps only its own charge \( CV \), so this working does not hold.
Working as per the standard method
After separation each capacitor keeps its charge \( \pm Q = \pm CV \) and its potential difference \( V \). The joined plates form an isolated conductor with net charge \( -Q + Q = 0 \), so no charge flows: charge is not destroyed (Option 1), and the free plates keep \( +Q \) and \( -Q \) (Option 2). In series the potential differences add:
\( \displaystyle V_{\mathrm{free}} = V + V = 2V \)
Hence, as per the standard method, the potential difference between the free plates is \( 2V \), and the correct answer is Option (3).
NCERT (2026-27):
Capacitors in Series – “… Thus, in the series combination, charges on the two plates (±Q) are the same on each capacitor. The total potential drop V across the combination is the sum of the potential drops V1 and V2 across C1 and C2, respectively.”