Discussion Forum

A capacitor of capacitance 4$\mu$ F  is charged to a potential difference of 6 V with a battery. The battery is removed and in its place another capacitor of capacitance is 8$\mu$ F introduced and the circuit is closed. The potential difference attained by each of the capacitors in V is 

Answer:

Explanation:

The final circuit  of the series capacitor system is shown in figure,

 Since , we know that in a series capacitors system, the charge stored by both the capacitors is equal. 

Hence , Q = $C_{eq} V$

 where,  $C_{eq}=\frac{4\times8}{4+8}=\frac{8}{3}\mu F\Rightarrow Q=\frac{8}{3}\times 6=16\mu C$

Hence, $V_{1}=\frac{Q}{C_{1}}=\frac{16}{4}=4V $   and   $V_{2}=\frac{Q}{C_{2}}=\frac{16}{8}=2V $

 Hence, the potential difference,

$\triangle V= V_{1}-V_{2}$

$\Rightarrow$    $\triangle$ V= 4-2=2V

Hence , the correct option is (a)