$2\,\mu F$ capacitance has potential difference across its two terminals $200\;volts$. It is disconnected with battery and then another uncharged capacitance is connected in parallel to it, then $P.D.$ becomes $20\;volts$. Then the capacity of another capacitance will be.......$\mu F$
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(c) By using, common potential $V = \frac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}$
$==>$ $20 = \frac{{2 \times 200 + {C_2} \times 0}}{{2 + {C_2}}}$ $==>$ ${C_2} = 18\,\mu F$
$==>$ $20 = \frac{{2 \times 200 + {C_2} \times 0}}{{2 + {C_2}}}$ $==>$ ${C_2} = 18\,\mu F$
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