There are two identical capacitors, the first one is uncharged and filled with a dielectric of constant K while the other one is charged to potential V having air between its plates. If two capacitors are joined end to end, the common potential will be

Question

There are two identical capacitors, the first one is uncharged and filled with a dielectric of constant K while the other one is charged to potential V having air between its plates. If two capacitors are joined end to end, the common potential will be (A) (V/K-1) (B) (K V/K+1) (C) (K V/K-1) (D) (V/K+1)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/314feffec802cda87d1a8e5851c19817-.png" /> V'=(K C(0)+C V/K C+C) V'=(C V/C(I+K)) V'=(V/l+K)

Q. There are two identical capacitors, the first one is uncharged and filled with a dielectric of constant $K$ while the other one is charged to potential $V$ having air between its plates. If two capacitors are joined end to end, the common potential will be

$\frac{V}{K - 1}$

$\frac{K V}{K + 1}$

$\frac{K V}{K - 1}$

$\frac{V}{K + 1}$

$V^{'} = \frac{K C ( 0 ) + C V}{K C + C}$
$V^{'} = \frac{C V}{C ( I + K )}$
$V^{'} = \frac{V}{l + K}$

Q. There are two identical capacitors, the first one is uncharged and filled with a dielectric of constant K while the other one is charged to potential V having air between its plates. If two capacitors are joined end to end, the common potential will be

K−1V​

K+1KV​

K−1KV​

K+1V​