A 2 μ F capacitor C 1 is first charged to a potential difference of 10 V using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor C 2 of 8 μ F. The charge in C 2 on equilibrium condition is μ C . (Round off to the Nearest Inteorer)

Question

A 2 μ F capacitor C 1 is first charged to a potential difference of 10 V using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor C 2 of 8 μ F. The charge in C 2 on equilibrium condition is μ C . (Round off to the Nearest Inteorer) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/b2f82d1f2d2bc2cba34635d39199d603-.png" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

20=( C 1+ C 2) V ⇒ V =2 volt. Q 2= C 2 V =16 μ C =16

Q. A 2μF capacitor C1​ is first charged to a potential difference of 10V using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor C2​ of 8μF. The charge in C2​ on equilibrium condition is_______ μC. (Round off to the Nearest Inteorer)

20=(C1​+C2​)V⇒V=2 volt. Q2​=C2​V=16μC =16

$20 = (C_{1} + C_{2}) V \Rightarrow V = 2$ volt.
$Q_{2} = C_{2} V = 16 μ C$
$= 16$