• # question_answer A capacitor of $6\mu F$ capacitance is charged up to 100 V. It is touched and then removed with another uncharged capacitor of $14\mu F,$ then ratio of charges on both and potential at $6\mu F$ capacitor is A)  6/14 and 50V    B)  3/7 and 30V C)  7/3 and 30V                       D)  None of these

Given,  ${{C}_{1}}=6\mu F,$ ${{C}_{2}}=14\mu F$ ${{V}_{1}}=100volt$ Charge on capacitor                 $q={{C}_{1}}{{V}_{1}}=6\times 100=600\mu C$ Redistribution of charge takes place in the ratio of their capacitances. $\therefore$  $\frac{{{q}_{1}}}{{{q}_{2}}}=\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{6}{14}=\frac{3}{7}$ Charge on capacitor ${{C}_{1}}$ after touching and then ${{C}_{1}}$removed,                 ${{q}_{1}}=\frac{3\times 600}{3+7}=180\mu C$ $\therefore$ Potential at $6\mu F$ capacitor