A) The potential difference of the condenser will become \[180\ volts\]
B) The P.D. will become \[20\ volts\]
C) The P.D. will remain unchanged
D) The charge on condenser will reduce to one third
Correct Answer: A
Solution :
For capacitor \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{d}_{1}}}{{{d}_{2}}}\]Þ\[{{V}_{2}}=\frac{{{V}_{1}}\times {{d}_{2}}}{{{d}_{1}}}=\frac{60\times 12}{4}=180\,V\]
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