Question

A capacitor of capacity $C_{1}$ is charged upto $V$ volt and then connected to an uncharged capacitor of capacity $C _{2}$. Then final potential difference across each will be

• A. $\frac{ C _{1} V _{1}+ C _{2} V _{2}}{ C _{1}+ C _{2}}$
• B. $\left(\frac{ C _{1}}{ C _{1}+ C _{2}}\right) V$
• C. $C_{1}+2 C_{2}$
• D. $\left(\frac{C_{1}+C_{2}}{C_{1} C_{2}}\right) V$

Answer 1

Answer: (B)

$C _{2}$ capacitor is uncharged capacitor that means potential difference across it is zero because capacity can't be zero
$C =\frac{\varepsilon_{0} A }{ d }, q = CV$
then potential difference will be
$V '=\frac{ C _{1} V _{1}+ C _{2} V _{2}}{ C _{1}+ C _{2}}$
$V _{2}=0, V _{1}= V$
then, $V '=\frac{ C _{1} V + C _{2}(0)}{ C _{1}+ C _{2}}$
$V '=\left(\frac{ C _{1}}{ C _{1}+ C _{2}}\right) V$