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Question

Physics Question on electrostatic potential and capacitance

nn identical capacitors are joined in parallel and charged to a potential VV . The charged capacitors are disconnected and then connected in series using insulating handles such that the positive plate of one is connected to the negative plate of the other. The potential difference across the free plates of the combination is

A

VV

B

V/n{V/n}

C

nVnV

D

(n+1)V(n+1)V

Answer

nVnV

Explanation

Solution

When nn identical capacitors are connected in parallel, each capacitor will charged to potential
VV and q=CVq=CV
Now, these capacitors are connected in series
\therefore Total potential difference across free plates of combination is
V=V1+V2+Vn=nVV'=V_{1}+V_{2}+ \dots V_{n}=nV